Everything, on one sheet

The card

"For all n in S" = AND over every case in S. Proving it true needs every case covered — by checking each (exhaustion) or a general argument covering them all at once; disproving it needs one false case.
Exhaustion (1.2): domain finite and bounded. List every case exactly, check each, state a conclusion — no separate mark for stating the bound itself.
General proof / contradiction (1.1): domain infinite or continuous, claim to be PROVED. Rearrange to a form true for every case at once (e.g. a squared term ⩾ 0), or assume it false and derive a contradiction.
Counter-example (1.3): claim to be DISPROVED, any domain size. Trial values, exhibit one failure with working shown, stop.
Bounded claim → exhaustion. Unbounded/continuous claim to prove → general algebra or contradiction. Any claim to disprove → counter-example.
Common losses: case list padded or short; conclusion omitted; product misread as sum; relation applied backwards; columns transposed; numerical examples treated as proof over an infinite domain; a divisibility claim asserted with no quotient shown.

Why it works — Why one counter-example is enough, and why exhaustion needs every case — the same fact, seen from both sides

A claim of the form "$f(n)$ is true for every $n$ in some set $S$" is not one statement — it is many statements joined together, one for each value of $n$ in $S$: $f(n_1)$ AND $f(n_2)$ AND $f(n_3)$ AND … , continuing for every element $S$ has. That is what "for every" or "for all" actually means: not a single claim about the set, but a claim about each member of it, all asserted at once. Two facts about AND now decide everything else in this topic. First, an AND is true only when every one of its parts is true — so if you want to prove the whole conjunction, there is no shortcut: you have to establish $f(n_1)$, and establish $f(n_2)$, and so on, for every single $n$ in $S$. If $S$ is finite and small enough to write out, that is exactly proof by exhaustion — list every case, check each one, and the conjunction is proved because you have directly verified every part of it. If $S$ is infinite, exhaustion is not slower, it is impossible: there is no way to finish writing out infinitely many checks, which is exactly why the spec's own exhaustion example restricts $x$ and $y$ to "odd integers less than 7" rather than leaving the set open-ended. Second, and this is the asymmetry that makes disproof cheap: an AND is false the moment even one of its parts is false, regardless of how many other parts are true. $f(n_1)$ AND $f(n_2)$ AND $f(n_3)$ is false if $f(n_2)$ alone is false, whatever $f(n_1)$ and $f(n_3)$ turn out to be. So disproving "$f(n)$ is true for every $n$" needs exactly one failing case, exhibited and checked — not because mathematicians are being lenient, but because one false part is structurally sufficient to make the whole AND false. This is also why the spec's counter-example guidance example is allowed to leave $n$ unbounded — "for all values of $n$" being an infinite conjunction is not a problem for disproof the way it is for proof, because disproof was never going to need to visit every part in the first place.

Traps — 9

algebraic-proof-attempted-when-exhaustion-required

The most-documented single error in this topic, confirmed independently in two separate series. October 2022: "a significant number of attempts... using logical or algebraic approaches. Such attempts were mixed in quality and often unnecessarily long-winded... Some who obtained a cubic expression for the product just concluded it was even without any justification." October 2023, on a different question: "an attempt at an algebraic solution which proved more demanding than that of a numerical one." When a claim is already restricted to a small, explicitly bounded set of cases, listing and checking them is both the method the spec names and — confirmed twice, independently — the faster and safer route.

numerical-examples-mistaken-for-general-proof

Confirmed as the single worst-performing question on its own paper: an October 2021 report on a "prove true for all positive x, y" question states "though only a third of candidates were able to score the first mark, there were many who attempted an algebraic proof... there was also a large number who considered that several numerical examples of the inequality holding constituted a proof" — 47% of candidates scored 0 out of 4. This is exhaustion's own logic reaching past where it applies: checking a handful of values and finding the claim true every time proves nothing about a domain with no last case to reach, however many values are checked. Only a general argument — one that holds for an arbitrary case, not a list of specific ones — closes an infinite or continuous domain; see the worked algebraic-proof example above for what that argument actually looks like.

multiple-of-k-assertion-without-quotient-shown

Confirmed on a real exhaustion question checking a "multiple of 4" claim — k > 2, so unlike "even" or "multiple of 5," the property is not visible by inspecting a last digit. The real mark scheme is explicit that a bare assertion does not earn the mark: "they can be divided by 4 (on its own) is insufficient without further clarification such as 'to give whole numbers'" (Oct 2023). What it does accept is a minimal but explicit marker that the division genuinely produces a whole number — a calculation such as "$40 \div 4 = 10$," a tick against each value, or the word "prove"/"QED" next to the working — anything that shows the check was actually done, not just declared. For a "multiple of 4" claim, write "$40 \div 4 = 10$, a whole number, so $40$ is a multiple of $4$," not "$40$ is divisible by $4$" on its own; the marker cannot tell a genuine check from a guess unless the quotient — or an equivalent explicit clarification — is shown.

product-misread-as-sum

Confirmed directly: "the definition of 'product' is not widely understood, with many students considering the sum of a, b and c instead" (Oct 2022). This is not an arithmetic mistake inside an otherwise correct method — it is checking the wrong quantity against every single case, so a perfectly executed exhaustive list still proves nothing about the question actually asked. Read what operation the question names before setting up the check, not after.

case-defining-relation-applied-backwards

Confirmed on a real exhaustion question where the cases were generated from a stated relationship between the variables: "sometimes the result of using c = b − 2 instead of c = b + 2" (Oct 2022). A relation like "c is 2 more than b" has a direction; reversing it silently generates a different — and wrong — set of cases from the very first line, so every case checked afterwards is checked against the wrong list, however carefully the rest of the working is done.

column-values-transposed-between-derived-variables

Confirmed on the same well-answered exhaustion question as above, as a separate, less prominent cause of lost marks: "a mix up between columns b and c" (Oct 2023). This is not the relation being applied backwards and not an arithmetic slip — each value can be individually correct, but written under the other variable's column head, so the row no longer records the case it claims to. It shows up specifically when two columns are both derived from the same base variable rather than from each other (here b and c are both computed from a, not from one another) — nothing forces the two labels apart the way a stated relation would, so check each entry against its own column head after filling a row in, not only that the pair of numbers is right somewhere in the row.

case-list-not-exactly-complete

Confirmed as a specific, separately-documented cause of lost marks, distinct from getting a case wrong: "the addition of extra rows" (Oct 2023) costs marks in one direction, and the mechanism above shows why an incomplete list costs them in the other — a bounded set has an exact size, established in the very first step of the proof, and the finished list has to match it exactly, neither padded with cases outside the stated bound nor missing any inside it.

conclusion-omitted

Confirmed on a well-answered exhaustion question, where the cases themselves were rarely the problem: "the solution merely required the sight of three correct rows and a minimal conclusion. Reasons for a loss of marks were:... the omission of a conclusion" (Oct 2023). A completed, correct case list is not yet a finished proof — the sentence stating that every case has been checked and the claim therefore holds is not decoration, it is the step that turns the list into a proof.

counter-example-search-continued-past-the-first-valid-one

Confirmed on a real counter-example question: "most candidates understood the principle of finding a counter-example, although many found far more than necessary" (June 2019) — the same report that states the underlying rule directly: "only one counter-example is required to prove that a statement is not true." Once one failing case has been found and justified, the disproof is complete; continuing to search costs time without earning further credit.

The card

"For all n in S" = AND over every case in S. Proving it true needs every case covered — by checking each (exhaustion) or a general argument covering them all at once; disproving it needs one false case.
Exhaustion (1.2): domain finite and bounded. List every case exactly, check each, state a conclusion — no separate mark for stating the bound itself.
General proof / contradiction (1.1): domain infinite or continuous, claim to be PROVED. Rearrange to a form true for every case at once (e.g. a squared term ⩾ 0), or assume it false and derive a contradiction.
Counter-example (1.3): claim to be DISPROVED, any domain size. Trial values, exhibit one failure with working shown, stop.
Bounded claim → exhaustion. Unbounded/continuous claim to prove → general algebra or contradiction. Any claim to disprove → counter-example.
Common losses: case list padded or short; conclusion omitted; product misread as sum; relation applied backwards; columns transposed; numerical examples treated as proof over an infinite domain; a divisibility claim asserted with no quotient shown.

Why it works — Why one counter-example is enough, and why exhaustion needs every case — the same fact, seen from both sides

A claim of the form "$f(n)$ is true for every $n$ in some set $S$" is not one statement — it is many statements joined together, one for each value of $n$ in $S$: $f(n_1)$ AND $f(n_2)$ AND $f(n_3)$ AND … , continuing for every element $S$ has. That is what "for every" or "for all" actually means: not a single claim about the set, but a claim about each member of it, all asserted at once. Two facts about AND now decide everything else in this topic. First, an AND is true only when every one of its parts is true — so if you want to prove the whole conjunction, there is no shortcut: you have to establish $f(n_1)$, and establish $f(n_2)$, and so on, for every single $n$ in $S$. If $S$ is finite and small enough to write out, that is exactly proof by exhaustion — list every case, check each one, and the conjunction is proved because you have directly verified every part of it. If $S$ is infinite, exhaustion is not slower, it is impossible: there is no way to finish writing out infinitely many checks, which is exactly why the spec's own exhaustion example restricts $x$ and $y$ to "odd integers less than 7" rather than leaving the set open-ended. Second, and this is the asymmetry that makes disproof cheap: an AND is false the moment even one of its parts is false, regardless of how many other parts are true. $f(n_1)$ AND $f(n_2)$ AND $f(n_3)$ is false if $f(n_2)$ alone is false, whatever $f(n_1)$ and $f(n_3)$ turn out to be. So disproving "$f(n)$ is true for every $n$" needs exactly one failing case, exhibited and checked — not because mathematicians are being lenient, but because one false part is structurally sufficient to make the whole AND false. This is also why the spec's counter-example guidance example is allowed to leave $n$ unbounded — "for all values of $n$" being an infinite conjunction is not a problem for disproof the way it is for proof, because disproof was never going to need to visit every part in the first place.

Traps — 9

algebraic-proof-attempted-when-exhaustion-required

The most-documented single error in this topic, confirmed independently in two separate series. October 2022: "a significant number of attempts... using logical or algebraic approaches. Such attempts were mixed in quality and often unnecessarily long-winded... Some who obtained a cubic expression for the product just concluded it was even without any justification." October 2023, on a different question: "an attempt at an algebraic solution which proved more demanding than that of a numerical one." When a claim is already restricted to a small, explicitly bounded set of cases, listing and checking them is both the method the spec names and — confirmed twice, independently — the faster and safer route.

numerical-examples-mistaken-for-general-proof

Confirmed as the single worst-performing question on its own paper: an October 2021 report on a "prove true for all positive x, y" question states "though only a third of candidates were able to score the first mark, there were many who attempted an algebraic proof... there was also a large number who considered that several numerical examples of the inequality holding constituted a proof" — 47% of candidates scored 0 out of 4. This is exhaustion's own logic reaching past where it applies: checking a handful of values and finding the claim true every time proves nothing about a domain with no last case to reach, however many values are checked. Only a general argument — one that holds for an arbitrary case, not a list of specific ones — closes an infinite or continuous domain; see the worked algebraic-proof example above for what that argument actually looks like.

multiple-of-k-assertion-without-quotient-shown

Confirmed on a real exhaustion question checking a "multiple of 4" claim — k > 2, so unlike "even" or "multiple of 5," the property is not visible by inspecting a last digit. The real mark scheme is explicit that a bare assertion does not earn the mark: "they can be divided by 4 (on its own) is insufficient without further clarification such as 'to give whole numbers'" (Oct 2023). What it does accept is a minimal but explicit marker that the division genuinely produces a whole number — a calculation such as "$40 \div 4 = 10$," a tick against each value, or the word "prove"/"QED" next to the working — anything that shows the check was actually done, not just declared. For a "multiple of 4" claim, write "$40 \div 4 = 10$, a whole number, so $40$ is a multiple of $4$," not "$40$ is divisible by $4$" on its own; the marker cannot tell a genuine check from a guess unless the quotient — or an equivalent explicit clarification — is shown.

product-misread-as-sum

Confirmed directly: "the definition of 'product' is not widely understood, with many students considering the sum of a, b and c instead" (Oct 2022). This is not an arithmetic mistake inside an otherwise correct method — it is checking the wrong quantity against every single case, so a perfectly executed exhaustive list still proves nothing about the question actually asked. Read what operation the question names before setting up the check, not after.

case-defining-relation-applied-backwards

Confirmed on a real exhaustion question where the cases were generated from a stated relationship between the variables: "sometimes the result of using c = b − 2 instead of c = b + 2" (Oct 2022). A relation like "c is 2 more than b" has a direction; reversing it silently generates a different — and wrong — set of cases from the very first line, so every case checked afterwards is checked against the wrong list, however carefully the rest of the working is done.

column-values-transposed-between-derived-variables

Confirmed on the same well-answered exhaustion question as above, as a separate, less prominent cause of lost marks: "a mix up between columns b and c" (Oct 2023). This is not the relation being applied backwards and not an arithmetic slip — each value can be individually correct, but written under the other variable's column head, so the row no longer records the case it claims to. It shows up specifically when two columns are both derived from the same base variable rather than from each other (here b and c are both computed from a, not from one another) — nothing forces the two labels apart the way a stated relation would, so check each entry against its own column head after filling a row in, not only that the pair of numbers is right somewhere in the row.

case-list-not-exactly-complete

Confirmed as a specific, separately-documented cause of lost marks, distinct from getting a case wrong: "the addition of extra rows" (Oct 2023) costs marks in one direction, and the mechanism above shows why an incomplete list costs them in the other — a bounded set has an exact size, established in the very first step of the proof, and the finished list has to match it exactly, neither padded with cases outside the stated bound nor missing any inside it.

conclusion-omitted

Confirmed on a well-answered exhaustion question, where the cases themselves were rarely the problem: "the solution merely required the sight of three correct rows and a minimal conclusion. Reasons for a loss of marks were:... the omission of a conclusion" (Oct 2023). A completed, correct case list is not yet a finished proof — the sentence stating that every case has been checked and the claim therefore holds is not decoration, it is the step that turns the list into a proof.

counter-example-search-continued-past-the-first-valid-one

Confirmed on a real counter-example question: "most candidates understood the principle of finding a counter-example, although many found far more than necessary" (June 2019) — the same report that states the underlying rule directly: "only one counter-example is required to prove that a statement is not true." Once one failing case has been found and justified, the disproof is complete; continuing to search costs time without earning further credit.

The card

"For all n in S" = AND over every case in S. Proving it true needs every case covered — by checking each (exhaustion) or a general argument covering them all at once; disproving it needs one false case.
Exhaustion (1.2): domain finite and bounded. List every case exactly, check each, state a conclusion — no separate mark for stating the bound itself.
General proof / contradiction (1.1): domain infinite or continuous, claim to be PROVED. Rearrange to a form true for every case at once (e.g. a squared term ⩾ 0), or assume it false and derive a contradiction.
Counter-example (1.3): claim to be DISPROVED, any domain size. Trial values, exhibit one failure with working shown, stop.
Bounded claim → exhaustion. Unbounded/continuous claim to prove → general algebra or contradiction. Any claim to disprove → counter-example.
Common losses: case list padded or short; conclusion omitted; product misread as sum; relation applied backwards; columns transposed; numerical examples treated as proof over an infinite domain; a divisibility claim asserted with no quotient shown.

Why it works — Why one counter-example is enough, and why exhaustion needs every case — the same fact, seen from both sides

A claim of the form "$f(n)$ is true for every $n$ in some set $S$" is not one statement — it is many statements joined together, one for each value of $n$ in $S$: $f(n_1)$ AND $f(n_2)$ AND $f(n_3)$ AND … , continuing for every element $S$ has. That is what "for every" or "for all" actually means: not a single claim about the set, but a claim about each member of it, all asserted at once. Two facts about AND now decide everything else in this topic. First, an AND is true only when every one of its parts is true — so if you want to prove the whole conjunction, there is no shortcut: you have to establish $f(n_1)$, and establish $f(n_2)$, and so on, for every single $n$ in $S$. If $S$ is finite and small enough to write out, that is exactly proof by exhaustion — list every case, check each one, and the conjunction is proved because you have directly verified every part of it. If $S$ is infinite, exhaustion is not slower, it is impossible: there is no way to finish writing out infinitely many checks, which is exactly why the spec's own exhaustion example restricts $x$ and $y$ to "odd integers less than 7" rather than leaving the set open-ended. Second, and this is the asymmetry that makes disproof cheap: an AND is false the moment even one of its parts is false, regardless of how many other parts are true. $f(n_1)$ AND $f(n_2)$ AND $f(n_3)$ is false if $f(n_2)$ alone is false, whatever $f(n_1)$ and $f(n_3)$ turn out to be. So disproving "$f(n)$ is true for every $n$" needs exactly one failing case, exhibited and checked — not because mathematicians are being lenient, but because one false part is structurally sufficient to make the whole AND false. This is also why the spec's counter-example guidance example is allowed to leave $n$ unbounded — "for all values of $n$" being an infinite conjunction is not a problem for disproof the way it is for proof, because disproof was never going to need to visit every part in the first place.

Traps — 9

algebraic-proof-attempted-when-exhaustion-required

The most-documented single error in this topic, confirmed independently in two separate series. October 2022: "a significant number of attempts... using logical or algebraic approaches. Such attempts were mixed in quality and often unnecessarily long-winded... Some who obtained a cubic expression for the product just concluded it was even without any justification." October 2023, on a different question: "an attempt at an algebraic solution which proved more demanding than that of a numerical one." When a claim is already restricted to a small, explicitly bounded set of cases, listing and checking them is both the method the spec names and — confirmed twice, independently — the faster and safer route.

numerical-examples-mistaken-for-general-proof

Confirmed as the single worst-performing question on its own paper: an October 2021 report on a "prove true for all positive x, y" question states "though only a third of candidates were able to score the first mark, there were many who attempted an algebraic proof... there was also a large number who considered that several numerical examples of the inequality holding constituted a proof" — 47% of candidates scored 0 out of 4. This is exhaustion's own logic reaching past where it applies: checking a handful of values and finding the claim true every time proves nothing about a domain with no last case to reach, however many values are checked. Only a general argument — one that holds for an arbitrary case, not a list of specific ones — closes an infinite or continuous domain; see the worked algebraic-proof example above for what that argument actually looks like.

multiple-of-k-assertion-without-quotient-shown

Confirmed on a real exhaustion question checking a "multiple of 4" claim — k > 2, so unlike "even" or "multiple of 5," the property is not visible by inspecting a last digit. The real mark scheme is explicit that a bare assertion does not earn the mark: "they can be divided by 4 (on its own) is insufficient without further clarification such as 'to give whole numbers'" (Oct 2023). What it does accept is a minimal but explicit marker that the division genuinely produces a whole number — a calculation such as "$40 \div 4 = 10$," a tick against each value, or the word "prove"/"QED" next to the working — anything that shows the check was actually done, not just declared. For a "multiple of 4" claim, write "$40 \div 4 = 10$, a whole number, so $40$ is a multiple of $4$," not "$40$ is divisible by $4$" on its own; the marker cannot tell a genuine check from a guess unless the quotient — or an equivalent explicit clarification — is shown.

product-misread-as-sum

Confirmed directly: "the definition of 'product' is not widely understood, with many students considering the sum of a, b and c instead" (Oct 2022). This is not an arithmetic mistake inside an otherwise correct method — it is checking the wrong quantity against every single case, so a perfectly executed exhaustive list still proves nothing about the question actually asked. Read what operation the question names before setting up the check, not after.

case-defining-relation-applied-backwards

Confirmed on a real exhaustion question where the cases were generated from a stated relationship between the variables: "sometimes the result of using c = b − 2 instead of c = b + 2" (Oct 2022). A relation like "c is 2 more than b" has a direction; reversing it silently generates a different — and wrong — set of cases from the very first line, so every case checked afterwards is checked against the wrong list, however carefully the rest of the working is done.

column-values-transposed-between-derived-variables

Confirmed on the same well-answered exhaustion question as above, as a separate, less prominent cause of lost marks: "a mix up between columns b and c" (Oct 2023). This is not the relation being applied backwards and not an arithmetic slip — each value can be individually correct, but written under the other variable's column head, so the row no longer records the case it claims to. It shows up specifically when two columns are both derived from the same base variable rather than from each other (here b and c are both computed from a, not from one another) — nothing forces the two labels apart the way a stated relation would, so check each entry against its own column head after filling a row in, not only that the pair of numbers is right somewhere in the row.

case-list-not-exactly-complete

Confirmed as a specific, separately-documented cause of lost marks, distinct from getting a case wrong: "the addition of extra rows" (Oct 2023) costs marks in one direction, and the mechanism above shows why an incomplete list costs them in the other — a bounded set has an exact size, established in the very first step of the proof, and the finished list has to match it exactly, neither padded with cases outside the stated bound nor missing any inside it.

conclusion-omitted

Confirmed on a well-answered exhaustion question, where the cases themselves were rarely the problem: "the solution merely required the sight of three correct rows and a minimal conclusion. Reasons for a loss of marks were:... the omission of a conclusion" (Oct 2023). A completed, correct case list is not yet a finished proof — the sentence stating that every case has been checked and the claim therefore holds is not decoration, it is the step that turns the list into a proof.

counter-example-search-continued-past-the-first-valid-one

Confirmed on a real counter-example question: "most candidates understood the principle of finding a counter-example, although many found far more than necessary" (June 2019) — the same report that states the underlying rule directly: "only one counter-example is required to prove that a statement is not true." Once one failing case has been found and justified, the disproof is complete; continuing to search costs time without earning further credit.

The card

Division identity: f(x) ≡ (ax − b)q(x) + r; r is constant for a linear divisor.
Remainder theorem: dividing by (ax − b) leaves f(b/a); by (ax + b), f(−b/a). Solve the bracket, never read it off.
Factor theorem: (ax − b) is a factor ⟺ f(b/a) = 0 — the same identity with r = 0.
Set f(root) equal to the stated remainder, not 0, unless told 'factor'.
Factorise completely: divide out the known factor, factorise the quadratic quotient, answer as a product of integer-coefficient linear factors.
Remainder wanted? Substitute. Quotient wanted? Divide, or compare coefficients.

Why it works — Where both theorems come from — one identity, evaluated at the value that deletes the quotient

Start with what division produces, before either theorem is stated. Dividing $f(x)$ by a linear expression $(ax - b)$ gives some quotient $q(x)$ and leaves some remainder $r$, and that is exactly the statement $$f(x) \equiv (ax - b)\,q(x) + r.$$ Two details in that line are doing all the work later, so they are worth pinning down now. First, $r$ is a constant. The division only stops when what is left over is too small to be divided again — if anything of degree 1 or more remained, the divisor (degree 1) would still go into it — so the remainder's degree must be strictly less than the divisor's, and "less than degree 1" means degree 0: a plain number, possibly zero. Second, the symbol is $\equiv$, not $=$. This is an identity: the two sides are the same expression written two ways, so it is true for every single value of $x$, not for a special few. That is not pedantry — it is the entire lever. Because the identity holds for every $x$, you are free to choose the most convenient $x$ there is. Choose the value that makes the divisor zero. $(ax - b) = 0$ when $x = \frac{b}{a}$, so substituting that value gives $$f\!\left(\tfrac{b}{a}\right) = \left(a \cdot \tfrac{b}{a} - b\right) q\!\left(\tfrac{b}{a}\right) + r = 0 \times q\!\left(\tfrac{b}{a}\right) + r = r.$$ That is the remainder theorem, and notice what it cost: nothing was assumed about $q(x)$ at all. Whatever the quotient is — however long, however awkward — it is multiplied by zero and disappears. This is why substitution beats division for a remainder: the division's entire output, the quotient, is the part you were never asked for, and choosing $x = \frac{b}{a}$ deletes it in one step rather than computing it and then throwing it away. The factor theorem is now not a second fact. Saying $(ax-b)$ is a factor of $f(x)$ means it divides exactly — i.e. $r = 0$ — and the identity then reads $f(x) \equiv (ax-b)q(x)$. Run the same substitution: $f\!\left(\frac{b}{a}\right) = r = 0$. And the converse runs backwards down the identical line: if $f\!\left(\frac{b}{a}\right) = 0$ then $r = 0$, so $f(x) \equiv (ax-b)q(x)$ and $(ax-b)$ is a factor. Both directions come from the one identity, which is why the factor theorem is an "if and only if" rather than a one-way test: $(ax - b)$ is a factor of $f(x)$ exactly when $f\!\left(\frac{b}{a}\right) = 0$. Everything the exam does to trip you here is a distortion of the substitution step, so read it once more slowly. The value substituted is the value that makes the *divisor* zero — solve $ax - b = 0$, do not read a number off the bracket. For $(x - 2)$ that value is $+2$; for $(x + 2)$ it is $-2$; for $(2x + 1)$ it is $-\frac{1}{2}$, not $\frac{1}{2}$ and not $-2$. And what the substitution equals is $r$, the *stated remainder* — which is zero only when the question has actually said "factor".

Traps — 8

root-sign-taken-straight-from-the-bracket

The dominant error on this spec item, confirmed on a real three-part factor-and-remainder question in both directions at once: candidates were recorded "setting $f(\frac{1}{2}) = 0$ in part (a) and/or setting $f(-2) = -25$ in part (b)" where the correct substitutions were $f(-\frac{1}{2})$ and $f(2)$ (Oct 2023, Q4). The fix is mechanical rather than mnemonic: never read a number off a bracket, solve the bracket for $x$. $(2x+1)=0$ gives $x = -\frac{1}{2}$; $(x+2)=0$ gives $x = -2$; $(3x-2)=0$ gives $x = \frac{2}{3}$. One extra line, and the most expensive error on the topic cannot happen.

remainder-equation-set-to-zero

Confirmed on an accessible opening question: the commonest conceptual errors were "using $f(-2)$ or $f(3)$ rather than $f(2)$ or setting equal to $-3$ or $0$, rather than $3$" (Jan 2024, Q1). Setting the substitution equal to zero is the factor-theorem reflex applied to a remainder question — and it is worth seeing that it is the same identity in both cases, $f(x) \equiv (ax-b)q(x) + r$, with $r$ being whatever the question said it was. Zero is a value of $r$, not the default.

long-division-chosen-when-substitution-was-asked-for

Not wrong, but measurably worse, and the examiners have said so in as many words: "Candidates who chose to use division gave themselves a more difficult task and tended to be less successful, with many giving up part way through the division" (Jan 2024, Q1), on a question where "the remainder theorem was the most efficient approach." Division computes the entire quotient in order to discard it. Reach for it when a later part actually wants the quotient; otherwise substitute.

factorisation-stopped-one-step-short

Confirmed as a specific, separately-recorded cause of lost marks: "leaving the answer to (c) as $(2x+1)(2x^2 + x - 15)$" (Oct 2023, Q4). That quadratic does factorise — $2x^2 + x - 15 = (2x-5)(x+3)$ — which is exactly why the answer was marked short: "factorise completely" is not satisfied while any factor still factorises. After dividing out the known factor, always ask whether the quadratic quotient breaks down further, and check its discriminant if it is not obvious.

hence-factorise-reached-without-using-the-given-factor

On a "hence factorise" part, the word "hence" is a mark-scheme instruction, not decoration: the working must derive the factorisation from the already-established factor, and reaching the identical, completely correct final answer by any other route earns nothing on that part. A real mark scheme is explicit about it, on the same question this lesson's worked chain and marked solution are modelled on: "The question states 'Hence' so there is an expectation that they use the factor of $(2x+1)$ which is given. Solutions that just state $4x^3 + 4x^2 - 29x - 15 = (2x+1)(2x-5)(x+3)$ score M0 and therefore A0 A0" (Oct 2023, Q4(c)(ii)) — a fully correct factorisation, worth zero, purely because the shown working never used the given factor. The matching real error is recorded in the same series' examiner report: candidates "using a calculator to solve $4x^3 + 4x^2 - 29x - 15 = 0$ and writing $x = -3, -\frac{1}{2}, \frac{5}{2}$" — solving the cubic directly, on a paper where calculators are permitted throughout, instead of dividing out the factor the question had already given them. Every worked chain, chain drill and marked solution in this lesson models the division route for exactly this reason: on a "hence" question, showing the division or comparison is not supporting style, it is the entire mark allocation for that part, independent of whether the final product is right.

non-integer-factor-form-given

A genuine "mathematically equivalent but marked wrong" trap, confirmed on a real question whose final step was a factorisation: a candidate "giving the factor as $x - \frac{7}{2}$... This is incorrect as integer values were required" (June 2025 Regional 01R, Q8). $\left(x - \frac{7}{2}\right)$ and $(2x-7)$ have the same root and are the same factor up to a constant, and the scheme still wants the integer-coefficient form. When a root comes out as a fraction $\frac{b}{a}$, write the factor as $(ax - b)$.

answer-left-as-a-list-not-a-product

A real mark scheme states the required form of a factorised cubic explicitly, requiring it "written out as a product of factors (not as a list) and isw" (Jan 2020). Listing the three factors, or listing the three roots, is not the same object as their product — the question asked for $f(x)$ factorised, so the answer is an expression equal to $f(x)$. The isw (ignore subsequent working) note is the encouraging half: once the product is correctly written down, later scribbling does not undo it.

show-that-steps-omitted

On a "show that" part, the printed answer is not the thing being marked — the route to it is. A real report records marks lost for "not showing the $= 0$ or intermediate lines when attempting to prove (a)" (Oct 2023, Q4), and the general marking guidance's cso convention makes the standard explicit: "There must be no errors in this part of the question to obtain this mark." Write the substitution, write the line that clears the fractions, then write the printed result — a solution that leaps from one to the other has demonstrated nothing that could be checked.

The card

(x−a)² + (y−b)² = r²: centre (a,b), radius r. Sign flips — centre coordinate is minus the value in the bracket.
General form x²+y²+2gx+2fy+c=0: complete the square first. Centre (−g,−f); r² = g²+f²−c.
Radius ⊥ tangent at any point on the circle: gradients multiply to −1.
Perpendicular from centre bisects a chord. Angle in a semicircle = 90° for any point on the circle.
Give radii and coordinates exactly (surd form) — never a rounded decimal unless told otherwise.

Why it works — Where the circle's equation actually comes from — Pythagoras, applied to a point that moves

Take a centre $(a, b)$ and ask which points $(x, y)$ sit at distance exactly $r$ from it. Draw the right-angled triangle formed by the horizontal gap $x - a$, the vertical gap $y - b$, and the straight-line distance between the two points as the hypotenuse. Pythagoras' theorem gives that distance directly: $\sqrt{(x-a)^2 + (y-b)^2} = r$. Squaring both sides — legal here because both sides are non-negative, a distance and a radius — removes the square root and produces exactly $(x-a)^2 + (y-b)^2 = r^2$. Nobody invented this equation as a definition to memorise; it is the distance formula, itself just Pythagoras' theorem restated in coordinates, applied to every point that happens to be exactly $r$ away from one fixed point. This is also why the coefficients of $x^2$ and $y^2$ in ANY valid circle equation are always equal (both $1$, after dividing through if needed): both terms come from squaring a genuine coordinate gap in the same Pythagorean sum, and an equation where they differ, or where an $xy$ term appears, is not describing a circle at all. Now multiply the standard form out and see what general form actually is. $(x-a)^2 + (y-b)^2 = r^2$ expands to $x^2 - 2ax + a^2 + y^2 - 2by + b^2 = r^2$, which rearranges to $x^2 + y^2 - 2ax - 2by + (a^2+b^2-r^2) = 0$. Matching this term by term against $x^2+y^2+2gx+2fy+c=0$ gives $2g = -2a$, so $a = -g$; identically $b = -f$; and $c = a^2+b^2-r^2$, so $r^2 = a^2+b^2-c = g^2+f^2-c$. Every part of "complete the square, then read off the centre and radius" is just running this exact expansion in reverse — the coefficient of $x$ is always $-2 \times$(the centre's x-coordinate), which is precisely why the centre's x-coordinate is minus half that coefficient, and the constant term always has $r^2$ hidden inside it as $g^2+f^2-c$, which is precisely why the radius cannot be read off without first isolating that combination. There is no separate fact to memorise here beyond this one piece of algebra, expanded once and then run the other way every time a general-form equation shows up.

Traps — 8

centre-sign-not-flipped

A real WMA12 centre/radius question confirms candidates stating the circle's centre with a sign flip — the value read straight off a completed-square bracket, instead of its negative (Jan 2023, Q6). $(x+3)^2$ is zero at $x=-3$, not $x=3$: the bracket has to be interpreted, not transcribed. This is the exact same mechanism the WMA11 pilot documents for a quadratic's turning point, and it costs a mark for the identical reason in both places.

general-form-read-as-centre-radius-form

Completing the square in both $x$ and $y$ is not optional preparation, it is the entire first step whenever a circle's equation is presented multiplied out — reading a centre or radius off general form directly, without converting first, is a genuine error type. It is not, however, the step most candidates actually lose the mark on: a real WMA12 general-form question confirms the vast majority of candidates successfully completed this conversion (Jan 2023, Q6), with the one error that does recur being the sign flip covered in the next trap item, not a failure to attempt the conversion at all.

solved-parameter-not-substituted-into-earlier-general-centre

When a circle's equation is given in terms of an unresolved constant (e.g. centre $(k, 2k)$, found correctly in an earlier part), and a later part requires a SPECIFIC numeric value of that constant — found by solving elsewhere in the same question — the mark for "identifying the centre" in that later part requires substituting the specific solved value back into the earlier general expression, not re-deriving from the original equation and not stopping after substituting the value into the circle's own equation without extracting actual coordinates. A real WMA12 examiner report confirms this directly, on a question where this exact substitution step was the single biggest mark-loss point of the whole question: "after nearly 90% of candidates successfully navigating the first two parts, fewer than 60% managed even the first mark, for identifying the centre of the circle... A few candidates chose the wrong value of k and so lost the mark, while others attempted to use the original equation to find where the circle meets the x axis thus making no progress at all with the question. Others simply put the value of k into the equation of the circle and then stopped. This did not gain any marks" (Jan 2021, Q9). Whenever a constant is left unresolved across multiple parts of the same question, the centre to use from that point on is the one built from the specific value just solved for — not the general symbolic form quoted correctly in an earlier part.

radius-given-as-decimal-not-exact-surd

The general marking guidance's exact-answer rule states plainly that "where, for example, an exact answer is asked for, or working with surds is clearly required, marks will normally be lost if the candidate resorts to using rounded decimals" — and a real WMA12 circle question specifically required an exact-form radius rather than a decimal approximation (Jan 2023, Q6). A radius of $2\sqrt{5}$ rounded to $4.47$ looks harmless; the mark scheme treats it as a lost mark.

implicit-differentiation-chosen-over-the-property

A tangent's gradient can, in principle, be found by implicitly differentiating the circle's equation instead of using the radius-perpendicularity property — but a real WMA12 report records that "very few attempted to differentiate the equation of the circle, and these often made little progress even though the next method marks were available" (Jan 2021, Q9). Valid on paper is not the same as safe under time pressure: the perpendicularity property reaches the same gradient in two lines, with no unfamiliar technique to execute correctly first.

origin-and-circle-centre-both-labelled-o

A real WMA12 question let a circle's centre and the point O — meaning the origin — sit in the same problem, and a report records "the misunderstanding that O was the centre of the circle rather than the origin... a special case was permitted, and candidates were able to score all but the final A mark" (Jan 2021, Q9). The scheme was generous that one time; reading which O is which before starting is the version of this habit that doesn't rely on generosity.

substitution-direction-not-checked-for-fractions

Before substituting a rearranged line into a circle equation, check which variable's coefficient is smaller — a real WMA12 mark scheme records that "the (−2y−10) substitution for x was more successful than the (−0.5x−5) for y which led to work involving fractions" on a genuine tangent-and-normal intersection question (Jan 2023, Q6). Both directions reach the same answer for the same marks; only one of them does it without an extra fraction-clearing step in the middle.

general-area-formula-chosen-over-axis-aligned-base-height

When a triangle's base lies along an axis, its height is simply the third vertex's coordinate on the other axis — no trig or formula needed. But a real WMA12 examiner report records that "many candidates [were] not using the simple formula for the area of the triangle, electing instead to use ½ab sinC" or even a shoelace formula, "not appreciating the geometry of the situation, with the requisite triangle being right-angled" (Jan 2021, Q9). The mark scheme itself credits either route as a complete method, so no mark is actually lost by reaching for the general formula — but the marked solution above finds T on the x-axis and reads the base OT and height (A's y-coordinate) straight off in two lines precisely because that axis-alignment is there to exploit, while ½ab sinC or a shoelace calculation forces extra side-length and angle work under time pressure for the same credit.

The card

uₙ = a + (n−1)d. Term n from term 1 takes (n−1) steps, not n.
Sₙ = ½n[2a+(n−1)d] = ½n(a+l), l = a+(n−1)d = last term. The (a+l) form needs no d.
Sum proof: write forwards, write reversed, add — every pair totals 2a+(n−1)d, n pairs.
Arithmetic: uₙ₊₁−uₙ = d, constant. Geometric (later): uₙ₊₁/uₙ = r, constant. Check which before choosing a formula.
Increasing: uₙ₊₁>uₙ ∀n. Decreasing: uₙ₊₁<uₙ ∀n. Periodic, period k: uₙ₊ₖ=uₙ ∀n.

Why it works — Why $u_n = a + (n-1)d$ — it falls out of the recurrence, it is not a separate rule to learn

Start from the recurrence itself, $u_{n+1} = u_n + d$, with first term $u_1 = a$, and apply it repeatedly rather than accepting the closed formula as given. $u_2 = u_1 + d = a + d$. $u_3 = u_2 + d = (a+d) + d = a + 2d$. $u_4 = u_3 + d = (a+2d) + d = a + 3d$. The pattern is exactly what "add $d$ every step" should produce: to reach $u_n$ from $u_1$ takes $(n-1)$ steps, each adding one more $d$, so $u_n = a + (n-1)d$. This is not a coincidence discovered by trying a few cases and guessing the pattern — it is a direct consequence of what "arithmetic" was defined to mean in the first place, and it is the same $(n-1)$-not-$n$ counting from the recurrence-relation teach block above, now packaged into a formula so that it never has to be counted by hand again. Two consequences follow directly. First, this is exactly why $u_1$ corresponds to $(n-1) = 0$ rather than $(n-1) = 1$: the first term needs zero applications of the recurrence to exist, and the formula has to agree with that or it would be wrong at its own starting point. Second, an arithmetic sequence is fully determined by exactly two numbers, $a$ and $d$ — every other term is a deterministic consequence of those two, which is the reason a question that gives you two arbitrary terms of an arithmetic sequence (not necessarily the first two) is always secretly a simultaneous-equations question: two unknowns, two given data points, one linear system.

Traps — 7

sum-proof-last-term-off-by-one

Confirmed directly on the standard sum-formula proof: "a surprising proportion of candidates struggled to write a correct expression of Sₙ at the start of their proof, often having the last term incorrect" (Jan 2020, Q8(i)). The fix is the same mechanism taught earlier in this lesson: reaching the nth term from the first takes (n − 1) steps of the recurrence, so the last term of an n-term series is a + (n − 1)d, never a + nd.

sum-proof-reversal-skipped

Confirmed on the same question: "Others failed to reverse the terms in the sum correctly, or at all" (Jan 2020, Q8(i)). Without the reversed copy, there is nothing for the forward copy to pair against, and the whole argument — every pair summing to the same constant — has no pairs to point at. This step is not optional decoration on the proof; it is the entire mechanism.

sum-proof-insufficient-terms-shown

Confirmed on the same question: candidates "did not display a sufficient number of terms to make subsequent steps in their proof to complete" (Jan 2020, Q8(i)). Two terms at each end plus an ellipsis — $a + (a+d) + \cdots + (a+(n-2)d) + (a+(n-1)d)$ — is the minimum that actually shows the pattern; a single "..." with no visible terms either side of it gives an examiner nothing to check the pairing against.

induction-attempted-for-standard-proof

Confirmed on the same question: "Attempts at proof by induction were also seen in a few cases, though little progress was made in most of these" (Jan 2020, Q8(i)). Induction can, in principle, prove this result, but it is not the proof the specification names, and the report records it as the less successful choice in practice. Where a question asks to "prove the standard formula," the pairing argument is the direct, expected route.

sum-proof-assumes-the-result

Confirmed on the same question: "Some candidates either started with, or used something that they were trying to prove, e.g. starting with $S_n = \frac{1}{2}n(a+L)$, or attempting to use a summation formula that is a generalisation of the result they needed to show" (Jan 2020, Q8(i)). This scores zero on an "ag" (answer given) proof no matter how cleanly the rest of the working is presented, because it assumes the very fact being proved rather than deriving it — a "prove that" question can never legitimately start from a form of its own target result. The only safe starting point is the definition of the series itself, written out term by term, exactly as the mechanism block above does.

sum-proof-pairing-fails-for-odd-n

Confirmed on the same question: some candidates instead tried pairing terms within a single copy of the sum — first-plus-last, second-plus-second-to-last, and so on — "but they did not give due consideration to what happens when there are an odd number of terms, and were unable to score the accuracy mark as a result" (Jan 2020, Q8(i)). An odd number of terms leaves one unpaired middle term this method has no clean way to handle, which is exactly why the write-forwards/write-backwards/add version taught in this lesson is the safer standard route — it pairs the full sum against a full reversed copy of itself, so it never needs $n$ to be even and never has a leftover term to account for separately.

arithmetic-and-geometric-formulas-confused

Confirmed as a recurring, named error category across at least two series (June 2019 and October 2020): candidates reach for the wrong series formula mid-question — applying the geometric sum formula to an arithmetic series, or the reverse — because the two topics sit close together and use similarly-shaped notation. The one-line check that catches it before it starts: an arithmetic sequence has a constant DIFFERENCE between consecutive terms ($u_{n+1} - u_n = d$, the same value every time); a geometric one has a constant RATIO instead. Confirm which quantity is actually constant before reaching for either sum formula.

The card

uₙ = a + (n−1)d. Term n from term 1 takes (n−1) steps, not n.
Sₙ = ½n[2a+(n−1)d] = ½n(a+l), l = a+(n−1)d = last term. The (a+l) form needs no d.
Sum proof: write forwards, write reversed, add — every pair totals 2a+(n−1)d, n pairs.
Arithmetic: uₙ₊₁−uₙ = d, constant. Geometric (later): uₙ₊₁/uₙ = r, constant. Check which before choosing a formula.
Increasing: uₙ₊₁>uₙ ∀n. Decreasing: uₙ₊₁<uₙ ∀n. Periodic, period k: uₙ₊ₖ=uₙ ∀n.

Why it works — Why $u_n = a + (n-1)d$ — it falls out of the recurrence, it is not a separate rule to learn

Start from the recurrence itself, $u_{n+1} = u_n + d$, with first term $u_1 = a$, and apply it repeatedly rather than accepting the closed formula as given. $u_2 = u_1 + d = a + d$. $u_3 = u_2 + d = (a+d) + d = a + 2d$. $u_4 = u_3 + d = (a+2d) + d = a + 3d$. The pattern is exactly what "add $d$ every step" should produce: to reach $u_n$ from $u_1$ takes $(n-1)$ steps, each adding one more $d$, so $u_n = a + (n-1)d$. This is not a coincidence discovered by trying a few cases and guessing the pattern — it is a direct consequence of what "arithmetic" was defined to mean in the first place, and it is the same $(n-1)$-not-$n$ counting from the recurrence-relation teach block above, now packaged into a formula so that it never has to be counted by hand again. Two consequences follow directly. First, this is exactly why $u_1$ corresponds to $(n-1) = 0$ rather than $(n-1) = 1$: the first term needs zero applications of the recurrence to exist, and the formula has to agree with that or it would be wrong at its own starting point. Second, an arithmetic sequence is fully determined by exactly two numbers, $a$ and $d$ — every other term is a deterministic consequence of those two, which is the reason a question that gives you two arbitrary terms of an arithmetic sequence (not necessarily the first two) is always secretly a simultaneous-equations question: two unknowns, two given data points, one linear system.

Traps — 7

sum-proof-last-term-off-by-one

Confirmed directly on the standard sum-formula proof: "a surprising proportion of candidates struggled to write a correct expression of Sₙ at the start of their proof, often having the last term incorrect" (Jan 2020, Q8(i)). The fix is the same mechanism taught earlier in this lesson: reaching the nth term from the first takes (n − 1) steps of the recurrence, so the last term of an n-term series is a + (n − 1)d, never a + nd.

sum-proof-reversal-skipped

Confirmed on the same question: "Others failed to reverse the terms in the sum correctly, or at all" (Jan 2020, Q8(i)). Without the reversed copy, there is nothing for the forward copy to pair against, and the whole argument — every pair summing to the same constant — has no pairs to point at. This step is not optional decoration on the proof; it is the entire mechanism.

sum-proof-insufficient-terms-shown

Confirmed on the same question: candidates "did not display a sufficient number of terms to make subsequent steps in their proof to complete" (Jan 2020, Q8(i)). Two terms at each end plus an ellipsis — $a + (a+d) + \cdots + (a+(n-2)d) + (a+(n-1)d)$ — is the minimum that actually shows the pattern; a single "..." with no visible terms either side of it gives an examiner nothing to check the pairing against.

induction-attempted-for-standard-proof

Confirmed on the same question: "Attempts at proof by induction were also seen in a few cases, though little progress was made in most of these" (Jan 2020, Q8(i)). Induction can, in principle, prove this result, but it is not the proof the specification names, and the report records it as the less successful choice in practice. Where a question asks to "prove the standard formula," the pairing argument is the direct, expected route.

sum-proof-assumes-the-result

Confirmed on the same question: "Some candidates either started with, or used something that they were trying to prove, e.g. starting with $S_n = \frac{1}{2}n(a+L)$, or attempting to use a summation formula that is a generalisation of the result they needed to show" (Jan 2020, Q8(i)). This scores zero on an "ag" (answer given) proof no matter how cleanly the rest of the working is presented, because it assumes the very fact being proved rather than deriving it — a "prove that" question can never legitimately start from a form of its own target result. The only safe starting point is the definition of the series itself, written out term by term, exactly as the mechanism block above does.

sum-proof-pairing-fails-for-odd-n

Confirmed on the same question: some candidates instead tried pairing terms within a single copy of the sum — first-plus-last, second-plus-second-to-last, and so on — "but they did not give due consideration to what happens when there are an odd number of terms, and were unable to score the accuracy mark as a result" (Jan 2020, Q8(i)). An odd number of terms leaves one unpaired middle term this method has no clean way to handle, which is exactly why the write-forwards/write-backwards/add version taught in this lesson is the safer standard route — it pairs the full sum against a full reversed copy of itself, so it never needs $n$ to be even and never has a leftover term to account for separately.

arithmetic-and-geometric-formulas-confused

Confirmed as a recurring, named error category across at least two series (June 2019 and October 2020): candidates reach for the wrong series formula mid-question — applying the geometric sum formula to an arithmetic series, or the reverse — because the two topics sit close together and use similarly-shaped notation. The one-line check that catches it before it starts: an arithmetic sequence has a constant DIFFERENCE between consecutive terms ($u_{n+1} - u_n = d$, the same value every time); a geometric one has a constant RATIO instead. Confirm which quantity is actually constant before reaching for either sum formula.

The card

uₙ = ar^(n−1); Sₙ = a(1−rⁿ)/(1−r) = a(rⁿ−1)/(r−1), r ≠ 1; S∞ = a/(1−r), |r| < 1 — all three printed in the booklet.
Proof: write Sₙ, write rSₙ, subtract — the shared middle terms cancel.
|r| < 1 because rⁿ is the only n-dependent part: |r| < 1 ⇒ rⁿ → 0.
Geometric: uₙ₊₁/uₙ constant. Arithmetic: uₙ₊₁ − uₙ constant. Check which first.
Mixed series: the shared term counts once — the GP starts at (shared term)×r.
Quote the formula before substituting; a bare calculator total scores nothing.

Why it works — Why $S_n = \frac{a(1-r^n)}{1-r}$ — multiply by $r$ and subtract, and why that is the move

The arithmetic sum proof works by writing the series a second time in reverse and adding, because arithmetic terms are related by *addition* — stepping forward through one copy and backward through the other changes each side by $+d$ and $-d$, which cancel, so every pair lands on the same total. None of that survives the switch to multiplication. Reverse a geometric series and add it to itself and the pairs are $a + ar^{n-1}$, $ar + ar^{n-2}$, $ar^2 + ar^{n-3}$ — try it with $3, 6, 12, 24$ and the pairs come out $27, 18, 18, 27$, plainly not constant. The arithmetic move fails here, and it fails for a reason you can name: constant pair-sums came from a constant additive step, and this series does not have one. What this series does have is a constant *multiplicative* step, so the move that exploits its structure is multiplication. Write the sum out in full: $S_n = a + ar + ar^2 + \cdots + ar^{n-2} + ar^{n-1}$ — and note the last term is $ar^{n-1}$, not $ar^n$, for the same $(n-1)$-counting reason established above. Now multiply the whole equation by $r$: $rS_n = ar + ar^2 + ar^3 + \cdots + ar^{n-1} + ar^n$ Look at what multiplying by $r$ actually did: it turned every term into the *next* term, which slides the entire list one place along. The second line is the first line shifted by one. That means the two lines share almost all of their terms — everything from $ar$ up to $ar^{n-1}$ appears in both — and only two terms fail to have a partner: the $a$ at the start of the first line, and the $ar^n$ at the end of the second. Subtracting is what collects that observation: $S_n - rS_n = a - ar^n$ Every shared term cancelled, exactly $n-1$ of them, because each appeared once on each side. Factorise both sides — $S_n(1-r) = a(1-r^n)$ — and divide by $(1-r)$ to finish: $S_n = \frac{a(1-r^n)}{1-r}$ That final division carries a condition worth stating rather than glossing: it is only legal when $1-r \neq 0$, i.e. $r \neq 1$. The formula genuinely does not apply when $r = 1$, and it does not need to — a geometric series with $r = 1$ is $a + a + a + \cdots$, the same number repeated, whose sum to $n$ terms is just $na$ by inspection. This is not an exception bolted onto the formula from outside; it is the one case the derivation itself excluded, showing up exactly where the derivation said it would. So the two proofs in this pair of lessons are not two unrelated tricks to memorise. Each one applies the operation that matches how its own series is built: an additively-generated series is attacked by adding a reversed copy, a multiplicatively-generated series is attacked by subtracting a multiplied copy. Knowing which is which is knowing what the series is.

Traps — 8

arithmetic-and-geometric-formulas-confused

Confirmed as a recurring, named error category across at least two series (June 2019 and October 2020): candidates reach for the wrong series type's formula mid-question, because the two topics sit adjacent in the specification, use similarly-shaped notation, and are printed one above the other on the same formula-booklet page. Seen from the geometric side, the tell is any appearance of $d$, of $(n-1)d$, or of $\frac{1}{2}n$ in working about a series whose terms are multiplying. The two-line check that catches it before a single mark is at stake: compute $u_{n+1} - u_n$ and $u_{n+1}/u_n$ for two consecutive pairs, and see which of the two comes out the same both times.

overlap-term-double-counted

Confirmed on a January 2022 mixed arithmetic-then-geometric question, verified as that paper's biggest discriminator. The examiner records that candidates "were not able to correctly process the overlap of the 20th term of the arithmetic series with the first term of the geometric series, and double counted this term" (Jan 2022). When one series ends on the same term another begins with, that term belongs to exactly one of the two sums. The report itself names both legal fixes: the geometric series starts at (shared term) × r, "or alternatively" the shared term is subtracted out of the arithmetic total — either one, never neither, never both.

geometric-series-restarted-from-the-original-first-term

Confirmed on the same January 2022 question, and distinct from the double-count above: "many candidates failed to understand the context with many assuming they had to start the geometric series from the first term of $a = 100$ again" (Jan 2022). Where a sequence changes character partway through, the geometric part inherits its first term from wherever the arithmetic part left off — not from the beginning of the whole sequence. Reading the question's "from the $k$th term onwards" literally, and writing down what that term actually equals before doing anything else, is the fix.

sum-to-infinity-first-term-misidentified

Confirmed on a January 2024 sum-to-infinity question: candidates "often used a = 6 instead of a = 1.5 in the sum to infinity formula" (Jan 2024, Q5(i)) — taking the sequence's original first term rather than the first term of the series actually being summed. This bites whenever a question asks for the sum of everything after some term, or re-expresses a series starting from a later index. The $a$ in $\frac{a}{1-r}$ is not "the first term of the sequence"; it is the first term of the sum you are being asked for. The self-check: the answer to "sum everything after the first term" must be smaller than the answer to "sum everything" by exactly that first term.

calculator-summation-shown-without-formula

Confirmed on the same January 2024 question: "a number resorted to using the summation button on their calculator which, even if [the value] was found, no marks were scored" (Jan 2024, Q5(i)). Calculators are permitted throughout this paper, which makes this trap easier to fall into rather than harder — a correct number with no visible formula and no visible substitution earns nothing, because the marks on offer are method marks and there is no method on the page to award them to. The mark scheme's standing principle says the same thing in general terms: where a learnt formula is used, "the formula should be quoted first."

standard-proofs-not-known

Confirmed on the January 2020 arithmetic sum-formula proof, with the examiner's advice naming both series explicitly: students "would be well advised to make sure they know the standard proofs for the arithmetic and geometric series to ensure they are able to give should it be required" (Jan 2020, Q8(i)). The specification's own guidance for item 4.4 says the same — "the proof of the sum formula for a finite series should be known." These are the only marks in the topic that a formula booklet cannot rescue: the formulae are printed, the proofs are not.

geometric-proof-subtraction-direction-reversed

Confirmed on the October 2020 geometric sum-formula proof: "A significant number started correctly and knew they had to subtract but subtracted the wrong way round to obtain e.g. $S_n - rS_n = ar^n - a$ and a significant number of candidates obtained $S_n - rS_n = a + ar^n$" (Oct 2020, Q8(a)). The correct line is $S_n - rS_n = a - ar^n$: the survivors of the subtraction are whichever terms belong to the line that was NOT shifted (the $a$ from $S_n$) minus whichever belong to the line that WAS (the $ar^n$ from $rS_n$), in that order — reverse which line each survivor is taken from, or turn the subtraction into an addition under exam pressure, and the result no longer factorises to the printed answer no matter how the rest of the algebra is handled. The examiner report records this immediately alongside the same question's other confirmed proof error (importing the arithmetic proof's reverse-and-add method), making the two the two most-recorded ways this specific proof goes wrong.

convergence-condition-read-as-r-less-than-one

Derived from the mechanism rather than quoted — the facts bank records no examiner finding on this specific error, and this entry carries no citation. Reading $|r| < 1$ as "$r < 1$" admits every negative ratio, including $r = -2$ and $r = -1.5$, whose terms grow without limit while alternating in sign. The modulus is a statement about size: $S_\infty$ exists precisely when each term is smaller in size than the one before it, regardless of sign. The related boundary case is $r = -1$, where $r^n$ alternates between $-1$ and $1$ forever and the partial sums alternate between $a$ and $0$ — never settling, which is why the specification's inequality is strict.

The card

uₙ = a + (n−1)d. Term n from term 1 takes (n−1) steps, not n.
Sₙ = ½n[2a+(n−1)d] = ½n(a+l), l = a+(n−1)d = last term. The (a+l) form needs no d.
Sum proof: write forwards, write reversed, add — every pair totals 2a+(n−1)d, n pairs.
Arithmetic: uₙ₊₁−uₙ = d, constant. Geometric (later): uₙ₊₁/uₙ = r, constant. Check which before choosing a formula.
Increasing: uₙ₊₁>uₙ ∀n. Decreasing: uₙ₊₁<uₙ ∀n. Periodic, period k: uₙ₊ₖ=uₙ ∀n.

Why it works — Why $u_n = a + (n-1)d$ — it falls out of the recurrence, it is not a separate rule to learn

Start from the recurrence itself, $u_{n+1} = u_n + d$, with first term $u_1 = a$, and apply it repeatedly rather than accepting the closed formula as given. $u_2 = u_1 + d = a + d$. $u_3 = u_2 + d = (a+d) + d = a + 2d$. $u_4 = u_3 + d = (a+2d) + d = a + 3d$. The pattern is exactly what "add $d$ every step" should produce: to reach $u_n$ from $u_1$ takes $(n-1)$ steps, each adding one more $d$, so $u_n = a + (n-1)d$. This is not a coincidence discovered by trying a few cases and guessing the pattern — it is a direct consequence of what "arithmetic" was defined to mean in the first place, and it is the same $(n-1)$-not-$n$ counting from the recurrence-relation teach block above, now packaged into a formula so that it never has to be counted by hand again. Two consequences follow directly. First, this is exactly why $u_1$ corresponds to $(n-1) = 0$ rather than $(n-1) = 1$: the first term needs zero applications of the recurrence to exist, and the formula has to agree with that or it would be wrong at its own starting point. Second, an arithmetic sequence is fully determined by exactly two numbers, $a$ and $d$ — every other term is a deterministic consequence of those two, which is the reason a question that gives you two arbitrary terms of an arithmetic sequence (not necessarily the first two) is always secretly a simultaneous-equations question: two unknowns, two given data points, one linear system.

Traps — 7

sum-proof-last-term-off-by-one

Confirmed directly on the standard sum-formula proof: "a surprising proportion of candidates struggled to write a correct expression of Sₙ at the start of their proof, often having the last term incorrect" (Jan 2020, Q8(i)). The fix is the same mechanism taught earlier in this lesson: reaching the nth term from the first takes (n − 1) steps of the recurrence, so the last term of an n-term series is a + (n − 1)d, never a + nd.

sum-proof-reversal-skipped

Confirmed on the same question: "Others failed to reverse the terms in the sum correctly, or at all" (Jan 2020, Q8(i)). Without the reversed copy, there is nothing for the forward copy to pair against, and the whole argument — every pair summing to the same constant — has no pairs to point at. This step is not optional decoration on the proof; it is the entire mechanism.

sum-proof-insufficient-terms-shown

Confirmed on the same question: candidates "did not display a sufficient number of terms to make subsequent steps in their proof to complete" (Jan 2020, Q8(i)). Two terms at each end plus an ellipsis — $a + (a+d) + \cdots + (a+(n-2)d) + (a+(n-1)d)$ — is the minimum that actually shows the pattern; a single "..." with no visible terms either side of it gives an examiner nothing to check the pairing against.

induction-attempted-for-standard-proof

Confirmed on the same question: "Attempts at proof by induction were also seen in a few cases, though little progress was made in most of these" (Jan 2020, Q8(i)). Induction can, in principle, prove this result, but it is not the proof the specification names, and the report records it as the less successful choice in practice. Where a question asks to "prove the standard formula," the pairing argument is the direct, expected route.

sum-proof-assumes-the-result

Confirmed on the same question: "Some candidates either started with, or used something that they were trying to prove, e.g. starting with $S_n = \frac{1}{2}n(a+L)$, or attempting to use a summation formula that is a generalisation of the result they needed to show" (Jan 2020, Q8(i)). This scores zero on an "ag" (answer given) proof no matter how cleanly the rest of the working is presented, because it assumes the very fact being proved rather than deriving it — a "prove that" question can never legitimately start from a form of its own target result. The only safe starting point is the definition of the series itself, written out term by term, exactly as the mechanism block above does.

sum-proof-pairing-fails-for-odd-n

Confirmed on the same question: some candidates instead tried pairing terms within a single copy of the sum — first-plus-last, second-plus-second-to-last, and so on — "but they did not give due consideration to what happens when there are an odd number of terms, and were unable to score the accuracy mark as a result" (Jan 2020, Q8(i)). An odd number of terms leaves one unpaired middle term this method has no clean way to handle, which is exactly why the write-forwards/write-backwards/add version taught in this lesson is the safer standard route — it pairs the full sum against a full reversed copy of itself, so it never needs $n$ to be even and never has a leftover term to account for separately.

arithmetic-and-geometric-formulas-confused

Confirmed as a recurring, named error category across at least two series (June 2019 and October 2020): candidates reach for the wrong series formula mid-question — applying the geometric sum formula to an arithmetic series, or the reverse — because the two topics sit close together and use similarly-shaped notation. The one-line check that catches it before it starts: an arithmetic sequence has a constant DIFFERENCE between consecutive terms ($u_{n+1} - u_n = d$, the same value every time); a geometric one has a constant RATIO instead. Confirm which quantity is actually constant before reaching for either sum formula.

The card

uₙ = ar^(n−1); Sₙ = a(1−rⁿ)/(1−r) = a(rⁿ−1)/(r−1), r ≠ 1; S∞ = a/(1−r), |r| < 1 — all three printed in the booklet.
Proof: write Sₙ, write rSₙ, subtract — the shared middle terms cancel.
|r| < 1 because rⁿ is the only n-dependent part: |r| < 1 ⇒ rⁿ → 0.
Geometric: uₙ₊₁/uₙ constant. Arithmetic: uₙ₊₁ − uₙ constant. Check which first.
Mixed series: the shared term counts once — the GP starts at (shared term)×r.
Quote the formula before substituting; a bare calculator total scores nothing.

Why it works — Why $S_n = \frac{a(1-r^n)}{1-r}$ — multiply by $r$ and subtract, and why that is the move

The arithmetic sum proof works by writing the series a second time in reverse and adding, because arithmetic terms are related by *addition* — stepping forward through one copy and backward through the other changes each side by $+d$ and $-d$, which cancel, so every pair lands on the same total. None of that survives the switch to multiplication. Reverse a geometric series and add it to itself and the pairs are $a + ar^{n-1}$, $ar + ar^{n-2}$, $ar^2 + ar^{n-3}$ — try it with $3, 6, 12, 24$ and the pairs come out $27, 18, 18, 27$, plainly not constant. The arithmetic move fails here, and it fails for a reason you can name: constant pair-sums came from a constant additive step, and this series does not have one. What this series does have is a constant *multiplicative* step, so the move that exploits its structure is multiplication. Write the sum out in full: $S_n = a + ar + ar^2 + \cdots + ar^{n-2} + ar^{n-1}$ — and note the last term is $ar^{n-1}$, not $ar^n$, for the same $(n-1)$-counting reason established above. Now multiply the whole equation by $r$: $rS_n = ar + ar^2 + ar^3 + \cdots + ar^{n-1} + ar^n$ Look at what multiplying by $r$ actually did: it turned every term into the *next* term, which slides the entire list one place along. The second line is the first line shifted by one. That means the two lines share almost all of their terms — everything from $ar$ up to $ar^{n-1}$ appears in both — and only two terms fail to have a partner: the $a$ at the start of the first line, and the $ar^n$ at the end of the second. Subtracting is what collects that observation: $S_n - rS_n = a - ar^n$ Every shared term cancelled, exactly $n-1$ of them, because each appeared once on each side. Factorise both sides — $S_n(1-r) = a(1-r^n)$ — and divide by $(1-r)$ to finish: $S_n = \frac{a(1-r^n)}{1-r}$ That final division carries a condition worth stating rather than glossing: it is only legal when $1-r \neq 0$, i.e. $r \neq 1$. The formula genuinely does not apply when $r = 1$, and it does not need to — a geometric series with $r = 1$ is $a + a + a + \cdots$, the same number repeated, whose sum to $n$ terms is just $na$ by inspection. This is not an exception bolted onto the formula from outside; it is the one case the derivation itself excluded, showing up exactly where the derivation said it would. So the two proofs in this pair of lessons are not two unrelated tricks to memorise. Each one applies the operation that matches how its own series is built: an additively-generated series is attacked by adding a reversed copy, a multiplicatively-generated series is attacked by subtracting a multiplied copy. Knowing which is which is knowing what the series is.

Traps — 8

arithmetic-and-geometric-formulas-confused

Confirmed as a recurring, named error category across at least two series (June 2019 and October 2020): candidates reach for the wrong series type's formula mid-question, because the two topics sit adjacent in the specification, use similarly-shaped notation, and are printed one above the other on the same formula-booklet page. Seen from the geometric side, the tell is any appearance of $d$, of $(n-1)d$, or of $\frac{1}{2}n$ in working about a series whose terms are multiplying. The two-line check that catches it before a single mark is at stake: compute $u_{n+1} - u_n$ and $u_{n+1}/u_n$ for two consecutive pairs, and see which of the two comes out the same both times.

overlap-term-double-counted

Confirmed on a January 2022 mixed arithmetic-then-geometric question, verified as that paper's biggest discriminator. The examiner records that candidates "were not able to correctly process the overlap of the 20th term of the arithmetic series with the first term of the geometric series, and double counted this term" (Jan 2022). When one series ends on the same term another begins with, that term belongs to exactly one of the two sums. The report itself names both legal fixes: the geometric series starts at (shared term) × r, "or alternatively" the shared term is subtracted out of the arithmetic total — either one, never neither, never both.

geometric-series-restarted-from-the-original-first-term

Confirmed on the same January 2022 question, and distinct from the double-count above: "many candidates failed to understand the context with many assuming they had to start the geometric series from the first term of $a = 100$ again" (Jan 2022). Where a sequence changes character partway through, the geometric part inherits its first term from wherever the arithmetic part left off — not from the beginning of the whole sequence. Reading the question's "from the $k$th term onwards" literally, and writing down what that term actually equals before doing anything else, is the fix.

sum-to-infinity-first-term-misidentified

Confirmed on a January 2024 sum-to-infinity question: candidates "often used a = 6 instead of a = 1.5 in the sum to infinity formula" (Jan 2024, Q5(i)) — taking the sequence's original first term rather than the first term of the series actually being summed. This bites whenever a question asks for the sum of everything after some term, or re-expresses a series starting from a later index. The $a$ in $\frac{a}{1-r}$ is not "the first term of the sequence"; it is the first term of the sum you are being asked for. The self-check: the answer to "sum everything after the first term" must be smaller than the answer to "sum everything" by exactly that first term.

calculator-summation-shown-without-formula

Confirmed on the same January 2024 question: "a number resorted to using the summation button on their calculator which, even if [the value] was found, no marks were scored" (Jan 2024, Q5(i)). Calculators are permitted throughout this paper, which makes this trap easier to fall into rather than harder — a correct number with no visible formula and no visible substitution earns nothing, because the marks on offer are method marks and there is no method on the page to award them to. The mark scheme's standing principle says the same thing in general terms: where a learnt formula is used, "the formula should be quoted first."

standard-proofs-not-known

Confirmed on the January 2020 arithmetic sum-formula proof, with the examiner's advice naming both series explicitly: students "would be well advised to make sure they know the standard proofs for the arithmetic and geometric series to ensure they are able to give should it be required" (Jan 2020, Q8(i)). The specification's own guidance for item 4.4 says the same — "the proof of the sum formula for a finite series should be known." These are the only marks in the topic that a formula booklet cannot rescue: the formulae are printed, the proofs are not.

geometric-proof-subtraction-direction-reversed

Confirmed on the October 2020 geometric sum-formula proof: "A significant number started correctly and knew they had to subtract but subtracted the wrong way round to obtain e.g. $S_n - rS_n = ar^n - a$ and a significant number of candidates obtained $S_n - rS_n = a + ar^n$" (Oct 2020, Q8(a)). The correct line is $S_n - rS_n = a - ar^n$: the survivors of the subtraction are whichever terms belong to the line that was NOT shifted (the $a$ from $S_n$) minus whichever belong to the line that WAS (the $ar^n$ from $rS_n$), in that order — reverse which line each survivor is taken from, or turn the subtraction into an addition under exam pressure, and the result no longer factorises to the printed answer no matter how the rest of the algebra is handled. The examiner report records this immediately alongside the same question's other confirmed proof error (importing the arithmetic proof's reverse-and-add method), making the two the two most-recorded ways this specific proof goes wrong.

convergence-condition-read-as-r-less-than-one

Derived from the mechanism rather than quoted — the facts bank records no examiner finding on this specific error, and this entry carries no citation. Reading $|r| < 1$ as "$r < 1$" admits every negative ratio, including $r = -2$ and $r = -1.5$, whose terms grow without limit while alternating in sign. The modulus is a statement about size: $S_\infty$ exists precisely when each term is smaller in size than the one before it, regardless of sign. The related boundary case is $r = -1$, where $r^n$ alternates between $-1$ and $1$ forever and the partial sums alternate between $a$ and $0$ — never settling, which is why the specification's inequality is strict.

The card

(a+bx)ⁿ, n a positive integer: n+1 terms, C(n,r)aⁿ⁻ʳ(bx)ʳ. Powers of a and bx in any term always add to n.
C(n,r) = n!/(r!(n−r)!) — a headcount of which r of n brackets give bx, not a formula to memorise alone.
The power of x in a term IS r. Write r = (power wanted) on its own line before anything else.
(bx)ʳ = bʳxʳ: raise the number too, not just the variable. Negative b alternates sign with the parity of r.
Pascal's row n has n+1 entries, C(n,0)…C(n,n). Check any answer: set x = 1, total should equal (a+b)ⁿ.
Writing several terms or a full Pascal's row? State the one answering the question on its own line — a correct value left unlabelled in a list can still lose the mark (Jan 2024 Q2).

Why it works — Where ⁿCᵣ actually comes from — it is a headcount, not a formula to memorise

Write $(a+bx)^n$ out as $n$ separate brackets multiplied together: $(a+bx)(a+bx)\cdots(a+bx)$. Expanding this product term by term means picking, from EVERY bracket, either the $a$ or the $bx$ — one choice per bracket — and multiplying all $n$ choices together; that is what "expanding a product of sums" means, mechanically. A term that ends up as $a^{n-r}(bx)^r$ is exactly the result of a selection that picked $bx$ from $r$ of the $n$ brackets and $a$ from the other $n - r$ — and every selection that picks the SAME set of $r$ brackets multiplies out to the identical term, so instead of writing that term out once per selection, the selections are counted and the count becomes the coefficient. $\binom{n}{r}$ is, by definition, the number of ways to choose which $r$ of $n$ brackets contribute the $bx$ — nothing more exotic than that, and nothing that needs a separate rule to justify why it's the coefficient. Now derive the factorial form from the same counting argument rather than quoting it: line the $n$ brackets up and decide, for each, whether it's an "$a$-bracket" or a "$bx$-bracket." There are $n!$ ways to put $n$ distinct objects in order — but the objects here aren't actually distinct once you only care about the FINAL split into two groups: swapping the order of the $r$ brackets already labelled "$bx$" among themselves doesn't create a new selection, so the $n!$ orderings over-count by a factor of $r!$ (the number of ways to shuffle those $r$ brackets among themselves); and swapping the order of the remaining $n-r$ "$a$"-brackets among themselves doesn't create a new selection either, over-counting by a further factor of $(n-r)!$. Dividing out both gives $\binom{n}{r} = \dfrac{n!}{r!(n-r)!}$ — not a formula handed down, but the direct consequence of removing the two kinds of reordering that don't actually change which brackets were chosen.

Traps — 6

term-index-mismatched-to-power

Confirmed directly on a real WMA12 question: common mistakes included "using the wrong power (usually 4 or 6 instead of 5 for the fractional part) or using the wrong binomial coefficient (¹²C₆, ¹²C₈ and ⁷C₅ instead of ¹²C₇)" (Jan 2024, Q2). The power of $x$ in the general term $\binom{n}{r}a^{n-r}(bx)^r$ is $r$ itself — no "one more" or "one less." State $r$ explicitly, on its own line, before touching any other number, and this error becomes almost impossible to make by accident.

general-series-form-chosen-over-combinatorial-form

The P2 formula booklet page prints two binomial formulas side by side: the combinatorial $(a+b)^n$ expansion for positive integer $n$, and the general series $(1+x)^n = 1+nx+\ldots$ for real $n$ and $|x|<1$ — the second is P4 content, not required here, and genuinely harder to apply. An examiner report confirms most candidates default correctly: "Part (a) was generally done very well with candidates mostly using the (a+b)ⁿ form of the binomial expansion where n is a positive integer rather than (1+x)ⁿ where n is a rational number" (June 2025, Q1) — but a regional variant paper shows exactly what the harder route costs once a question adds a second unknown: "Many candidates struggled to make a start with this question due to the fact that there were two unknowns, n and k, within the binomial expression (1+kx)ⁿ" (June 2025 Regional 01R, Q7). On P2, $n$ is always a known positive integer — default to $(a+b)^n$ every time.

bx-term-not-raised-as-a-whole

Confirmed directly on a real WMA12 question, not just derived from the arithmetic: the examiner report for the same anchor question cited elsewhere in this lesson names this as the most common way of losing the accuracy mark for the unsimplified term — "using invisible brackets that were not recovered in further work (4x⁷ instead of (4x)⁷) resulting in multiplication by 4 instead of 4⁷" (Jan 2024, Q2). $(bx)^r$ means both $b$ and $x$ are raised to the power $r$ — $(4x)^3 = 4^3x^3 = 64x^3$, not $4x^3$. Losing the power on the numeral while keeping it on the variable is easy to miss on the page, because the $x^3$ still looks correct; only the coefficient in front of it is wrong. Write $(bx)^r$ as $b^r x^r$ before evaluating $b^r$, so the two pieces are never handled in the same breath.

sign-lost-on-odd-powers-of-a-negative-b

When $b$ is negative, $(bx)^r = b^r x^r$ carries the sign of $b^r$ — negative when $r$ is odd, positive when $r$ is even, for the same reason $(-1)^3 = -1$ while $(-1)^4 = 1$. The terms of an expansion like $(3-2x)^7$ therefore alternate strictly by the parity of $r$, not by any other pattern — dropping the sign on just one term, usually by writing $(-2)^r$ as $2^r$ out of habit, produces an answer that looks structurally right and is numerically wrong by exactly a factor of $-1$.

pascal-triangle-row-miscounted

Row $n$ of Pascal's triangle — counting the single "1" at the top as row 0 — has exactly $n+1$ entries, $\binom{n}{0}$ through $\binom{n}{n}$. Starting the count at row 1 instead of row 0 reads off the coefficients for $n-1$, one whole row too early, and every coefficient taken from it is wrong even though every addition inside the triangle was done correctly. Before reading off a row, count its entries: $n+1$ numbers, not $n$.

required-term-not-isolated-from-a-list

Confirmed directly on the anchor WMA12 question: the examiner report states the final accuracy mark "was commonly not awarded to candidates who failed to isolate the required term from a list" (Jan 2024, Q2), and the mark scheme itself warns "the term or coefficient must be identified if they have more than one term." Writing several terms, a full expansion, or a whole row of Pascal's triangle is not itself an error — but once the value asked for has been found, it has to be pointed to on its own, not left sitting unlabelled inside the longer list. State it on its own final line (or circle/box it) the moment it's reached.

The card

logₐb = c ⇔ aᶜ = b (a>0, a≠1, b>0). y=aˣ passes through (0,1), asymptote y=0, never negative.
logₐ(xy)=logₐx+logₐy; logₐ(x/y)=logₐx−logₐy; logₐ(xᵏ)=k·logₐx; logₐ(1/x)=−logₐx; logₐa=1.
Change of base (in the booklet): logₐx = log_b x / log_b a. The other four laws are not — memorise them.
Solving aˣ=b: isolate the exponential first, then take logs — finish the rearrangement, not just the bracket.
If a question names the required form (e.g. x = logₐk), an unsimplified logₐb − c earns only the intermediate mark — combine it into the single required log to earn the final one (Jan 2021, Q3(i)).
A log of a sum does not split apart. Subtracting logs never divides the logs themselves, only their arguments.
A combined-log equation can gain a root the original excludes — check each argument is positive before accepting it.

Why it works — Why the laws hold — the laws of indices, read through one definition

Let $\log_a x = p$ and $\log_a y = q$ — by the definition above, this means precisely that $a^p = x$ and $a^q = y$, nothing more. Every law of logarithms is what happens to that pair of equations when you multiply, divide or raise them to a power, translated back into log notation. Multiply the two equations: $x \times y = a^p \times a^q$. The laws of indices already say $a^p \times a^q = a^{p+q}$ — indices add when you multiply powers of the same base — so $xy = a^{p+q}$. Taking $\log_a$ of both sides (asking 'what power of $a$ gives this?') recovers $p + q$ exactly, because $a^{p+q}$ passed back through $\log_a$ just returns its own exponent: $\log_a(xy) = p + q = \log_a x + \log_a y$. The subtraction law is the same argument run on division: $x/y = a^p/a^q = a^{p-q}$ by the quotient rule for indices, so $\log_a(x/y) = p - q = \log_a x - \log_a y$. The power law comes from raising the first equation to the power $k$: $x^k = (a^p)^k = a^{pk}$ by the laws of indices (a power raised to a power multiplies the exponents), so $\log_a(x^k) = pk = k\log_a x$ — the exponent $k$ moves from inside the log's argument to a multiplier in front, which is the entire content of the law, and the reason $(\log_a y)^2$ is a genuinely different expression from $\log_a(y^2)$: squaring the NUMBER $\log_a y$ is not the same operation as squaring the ARGUMENT $y$ before taking its log. The remaining two laws are special cases already covered: $\log_a(1/x) = \log_a 1 - \log_a x$ by the subtraction law, and $\log_a 1 = 0$ always (since $a^0 = 1$ for any valid base), so $\log_a(1/x) = -\log_a x$; and $\log_a a = 1$ is immediate from the definition itself, since $a^1 = a$. None of these five statements is a separate fact to memorise — they are the laws of indices, read backwards through the one definition that connects a power to a log.

Traps — 6

log-subtraction-rewritten-as-division

Confirmed directly on a real WMA12 log-manipulation question: "A fairly common mistake was to rewrite log₃(5x+7) − log₃4x as log₃(5x+7)/log₃4x. Some then crossed out log₃ in the numerator and denominator as though simplifying a fraction" (Jan 2023, Q4(i)). The subtraction law says $\log_a x - \log_a y = \log_a(x/y)$ — the two ARGUMENTS are divided, the two log SYMBOLS never are. Writing $\log_3(5x+7)/\log_34x$ treats "log₃" as a common factor that can be cancelled, which it is not: $\log_a$ is a function name, not a multiplier.

log-of-a-sum-expanded-as-sum-of-logs

Confirmed on the same question: "Another mistake, common for this type of question, was to expand the brackets within the log function, writing log₃(5x+7) as log₃5x + log₃7" (Jan 2023, Q4(i)). No law distributes a log across a sum or difference inside its own argument — the addition law only COMBINES two separate logs into one, it never SPLITS a single log open. $\log_a(x+y)$ has no simpler form; treating it like $\log_a(xy)$ applies the multiplication law to an expression that is added, not multiplied.

log-squared-confused-with-log-of-the-square

Confirmed on a harder question, described as "one of the most challenging parts on the paper": "understanding the difference between logₐ(y²) and (logₐy)²... rewriting (logₐy)² as 2logₐy was a very common mistake" (Jan 2023, Q4(ii)). The power law moves an exponent sitting INSIDE the log — it says nothing about an exponent sitting OUTSIDE the whole expression. $(\log_a y)^2$ means "square the number $\log_a y$ produces," and no law turns that squaring into a multiplication by 2 — the two expressions are built from the same symbols in a different order and are not generally equal.

exponent-rearrangement-left-unfinished

Confirmed on a solving-aˣ=b question with a measured mark-by-mark drop-off: "A fall off from nearly 90% to just over 70% between the first two marks is revealing" (Jan 2021, Q3(i)) — the gap between the M1 for setting up $(x+2)\log 7 = \log 3$ and the first A1 for reaching ANY correct exact expression for $x$. Taking logs correctly is not the same as finishing the algebra — once the exponent has been isolated as an expression like $2x-1$ or $x+1$, every remaining operation (adding, subtracting, dividing) still has to be carried out on both sides before an exact expression for $x$ exists at all. (A separate, later-stage error — reaching a correct exact expression for $x$ but leaving it uncombined into a single required log form — is a distinct trap: see "single-log-form-left-uncombined" below.)

single-log-form-left-uncombined

Confirmed on the same question, describing the steeper of its two mark-by-mark drops — "with only 50% scoring all three. For example, [many] achieved x = log₇3 − 2 but then did not reach the required form, failing to deal with the −2" (Jan 2021, Q3(i)). This is NOT an unfinished-rearrangement error: $x = \log_7 3 - 2$ is already a fully isolated, fully-credited exact expression for $x$ — it earns the A1 the trap above describes. The mark actually lost here is a further, distinct accuracy mark for combining that expression into the single logarithm the question specifically demanded ("giving your answer in the form $x = \log_7 a$, $a$ rational"): $\log_7 3 - 2 = \log_7 3 - \log_7 49 = \log_7\dfrac{3}{49}$, using $\log_7 49 = 2$ since $7^2 = 49$ — i.e. rewriting the constant $2$ via $\log_a a = 1$, here $2 = 2\log_7 7 = \log_7(7^2) = \log_7 49$. Whenever a question specifies the answer form this precisely, a correct but unsimplified expression like $\log_a b - c$ earns only the intermediate accuracy mark, never the final one.

logs-taken-before-the-equation-is-rearranged

A direct piece of technique advice from the same report, given as general guidance rather than a specific error: "Although the question asked for an answer in log form, it is not always advisable to take logs first when the equation is presented in index form" (Jan 2021, Q3(i)). Where a constant multiplies or divides the exponential term, isolating it algebraically first — before any logarithm is taken — removes an entire log law (usually the addition or subtraction law) from the working. Taking logs immediately is not wrong, but it adds a step, and every added step is an added chance for one of the other traps on this list to strike.

The card

logₐb = c ⇔ aᶜ = b (a>0, a≠1, b>0). y=aˣ passes through (0,1), asymptote y=0, never negative.
logₐ(xy)=logₐx+logₐy; logₐ(x/y)=logₐx−logₐy; logₐ(xᵏ)=k·logₐx; logₐ(1/x)=−logₐx; logₐa=1.
Change of base (in the booklet): logₐx = log_b x / log_b a. The other four laws are not — memorise them.
Solving aˣ=b: isolate the exponential first, then take logs — finish the rearrangement, not just the bracket.
If a question names the required form (e.g. x = logₐk), an unsimplified logₐb − c earns only the intermediate mark — combine it into the single required log to earn the final one (Jan 2021, Q3(i)).
A log of a sum does not split apart. Subtracting logs never divides the logs themselves, only their arguments.
A combined-log equation can gain a root the original excludes — check each argument is positive before accepting it.

Why it works — Why the laws hold — the laws of indices, read through one definition

Let $\log_a x = p$ and $\log_a y = q$ — by the definition above, this means precisely that $a^p = x$ and $a^q = y$, nothing more. Every law of logarithms is what happens to that pair of equations when you multiply, divide or raise them to a power, translated back into log notation. Multiply the two equations: $x \times y = a^p \times a^q$. The laws of indices already say $a^p \times a^q = a^{p+q}$ — indices add when you multiply powers of the same base — so $xy = a^{p+q}$. Taking $\log_a$ of both sides (asking 'what power of $a$ gives this?') recovers $p + q$ exactly, because $a^{p+q}$ passed back through $\log_a$ just returns its own exponent: $\log_a(xy) = p + q = \log_a x + \log_a y$. The subtraction law is the same argument run on division: $x/y = a^p/a^q = a^{p-q}$ by the quotient rule for indices, so $\log_a(x/y) = p - q = \log_a x - \log_a y$. The power law comes from raising the first equation to the power $k$: $x^k = (a^p)^k = a^{pk}$ by the laws of indices (a power raised to a power multiplies the exponents), so $\log_a(x^k) = pk = k\log_a x$ — the exponent $k$ moves from inside the log's argument to a multiplier in front, which is the entire content of the law, and the reason $(\log_a y)^2$ is a genuinely different expression from $\log_a(y^2)$: squaring the NUMBER $\log_a y$ is not the same operation as squaring the ARGUMENT $y$ before taking its log. The remaining two laws are special cases already covered: $\log_a(1/x) = \log_a 1 - \log_a x$ by the subtraction law, and $\log_a 1 = 0$ always (since $a^0 = 1$ for any valid base), so $\log_a(1/x) = -\log_a x$; and $\log_a a = 1$ is immediate from the definition itself, since $a^1 = a$. None of these five statements is a separate fact to memorise — they are the laws of indices, read backwards through the one definition that connects a power to a log.