Algebraic Proof

Topic

Course Support Current PowerPoint version

More questions on this topic: AS Pure: Proof (34) A2 Pure: Proof (107)

June 2025 Paper 2 Q8

AQACurrent spec3 marksAlgebraic Proof Expanding Brackets

8 \(k\) is a positive integer.

Prove that \(\quad 4k(k + 8) - (4k + 9)(k - 1) \quad\) is a multiple of 9 [3 marks]

Mark scheme

Answer	Mark	Comments
\(4k^2 + 32k\) or \(4k^2 - 4k + 9k - 9\) or \(4k^2 + 5k - 9\)	M1	may be seen in a full expansion eg \(\ldots - 4k^2 + 4k - 9k + 9\) may be seen in a grid
\(4k^2 + 32k - 4k^2 + 4k - 9k + 9\) or \(4k^2 + 32k - (4k^2 - 4k + 9k - 9)\) or \(4k^2 + 32k - 4k^2 - 5k + 9\) or \(4k^2 + 32k - (4k^2 + 5k - 9)\)	M1dep	correct 6-term expression and no errors or correct 5-term expression and no errors
M2 awarded and \(27k + 9\) and valid explanation involving 9 and \((3k + 1)\)	A1	eg M2 awarded and \(27k + 9 = 9(3k + 1)\) or M2 awarded and \(\dfrac{27k + 9}{9} = 3k + 1\)

Additional guidance

\(4k^2 - 4k + 9k - 9\) followed by an error in simplifying to 3 terms eg \(4k^2 - 4k + 9k - 9 = 4k^2 - 5k - 9 \quad (-5k\) is an error) Award 1st M1 but no further marks can be awarded	M1M0A0
Do not award 2nd M1 if errors seen eg \(4k(k + 8) = 4k^2 + 32k\) \((4k + 9)(k - 1) = 4k^2 + 4k - 9k - 9 = 4k^2 + 5k - 9 \quad (+4k\) and \(-9k\) are errors) \(4k^2 + 32k - 4k^2 - 5k + 9\) (‘correct’ 5-term expression with errors)	M1 M0A0
Missing brackets cannot be recovered
For M marks accept eg \(- -9\) for \(+9\)
Valid explanations involving 9 and \((3k + 1)\) include eg1 \(27k + 9\) when divided by 9 is \((3k + 1)\) eg2 \(27k + 9\) has factors \((3k + 1)\) and 9
Incomplete explanations include eg1 \(27k + 9\) which is divisible by 9 eg2 \(27k + 9\) and \(27k \div 9 = 3k\) and \(9 \div 9 = 1\)
Condone eg \(x\) used for \(k\)

June 2023 Paper 1 Q5

AQACurrent spec3 marksAlgebraic Proof Sequences

5 \(X_n\) and \(Y_n\) are the \(n\)th terms of two sequences.

\[X_n = (n - 1)(n + 1)\]\[Y_n = (n + 1)(n + 2)\]

Prove that every term of the sequence with \(n\)th term \(\quad Y_n - X_n \quad\) is a multiple of 3 [3 marks]

Mark scheme

Answer	Mark	Comments
\((n + 1)(.........)\) or \(n^2 + n - n - 1\) or \(n^2 - 1\) or \(n^2 + 2n + n + 2\) or \(n^2 + 3n + 2\)	M1	ie common factor of \((n + 1)\) taken out and attempt to subtract oe
\((n + 1)(n + 2 - (n - 1))\) or \(n^2 + 2n + n + 2 - (n^2 - n + n - 1)\)	M1dep	oe eg could be subtracted vertically
\((n + 1)(n + 2 - n + 1) = 3(n + 1)\) or \(3n + 3\) or \(n^2 + 3n + 2 - n^2 + 1 = 3n + 3\) or \(3(n + 1)\)	A1	no statement needed at end of proof

Additional guidance

\(X_n - Y_n\) can be awarded method marks but A0 unless rescued at the end eg recovery \(-3(n + 1)\) so \(Y_n - X_n\) is a multiple of 3	M2 A0 M2 A1
Missing brackets can be recovered later in the proof (but if not recovered then second M should not be awarded). Must see \(3n + 3\) to show recovery eg \(n^2 + 3n + 3 - n^2 - 1\) (error in \(X_n\) but \(Y_n\) correct with missing bracket)	M2 A1 M1 M0 A0
Condone use of \(x\) (or other letter) instead of \(n\). A mixture of letters would only gain M1. However \(Y_n - X_n = 3(n + 1) = 3x\) would be acceptable to show a multiple of 3

June 2022 Paper 2 Q6

AQACurrent spec3 marksAlgebraic Proof Expanding Brackets

6 Prove algebraically that when \(n\) is an integer

\[\frac{(2n + 1)^2 - (2n - 1)^2}{4} \quad \text{is always even.}\]

[3 marks]

Mark scheme

Answer	Mark	Comments
Alternative method 1 Expands the given brackets
\(((2n + 1)^2 =)\; 4n^2 + 2n + 2n + 1\) or \(((2n - 1)^2 =)\; 4n^2 - 2n - 2n + 1\)	M1	oe expansion eg \(((2n + 1)^2 =)\; 4n^2 + 4n + 1\) may be seen in a grid may be seen embedded in second mark ignore any denominator
\(4n^2 + 2n + 2n + 1 - 4n^2 + 2n + 2n - 1\) or \(4n^2 + 4n + 1 - 4n^2 + 4n - 1\) or \(4n^2 + 2n + 2n + 1 - (4n^2 - 2n - 2n + 1)\) and \(8n\) with no errors seen or \(4n^2 + 4n + 1 - (4n^2 - 4n + 1)\) and \(8n\) with no errors seen	M1dep	terms in any order ignore any denominator
M2 seen and valid explanation	A1	eg1 M2 seen and \(\dfrac{8n}{4} = 2n\) eg2 M2 seen and \(8n\) is even and when divided by 4 it is even
Alternative method 2 Difference of two squares
\((2n + 1 + 2n - 1)(2n + 1 - (2n - 1))\) or \((2n + 1 + 2n - 1)(2n + 1 - 2n + 1)\)	M1	ignore any denominator
M1 seen and \(4n \times 2\) with no errors seen	M1dep	ignore any denominator
M2 seen and valid explanation	A1	eg1 M2 seen and \(\dfrac{4n \times 2}{4} = 2n\) eg2 M2 seen and \(\dfrac{8n}{4} = 2n\) eg3 M2 seen and \(8n\) is even and when divided by 4 it is even

Additional guidance

Do not allow missing brackets even if recovered
Alt 1 \(4n^2 + 4n + 1 - 4n^2 - 4n + 1\)	M1M0
Alt 1 \(4n^2 + 2n + 2n + 1 - (4n^2 - 2n - 2n + 1)\) \(= 4n^2 + 4n + 1 - 4n^2 - 4n - 1 = 8n\) (\(8n\) but error seen)	M1 M0
Alt 1 Only \(8n\)	M0M0
Alt 1 2nd M1 Allow unnecessary brackets eg \(4n^2 + 4n + 1 - (4n^2 - 4n + 1) = (4n^2 - 4n^2) + (4n + 4n) + (1 - 1)\)	M1M1
Alt 2 \((2n + 1 + 2n - 1)(2n + 1 - 2n - 1)\)	M0M0
Alt 2 \((2n + 1 + 2n - 1)(2n + 1 - (2n - 1))\) \(= (2n + 1 + 2n - 1)(2n + 1 - 2n - 1) = 4n \times 2\) (\(4n \times 2\) but error seen)	M1 M0
Alt 2 \((2n + 1 + 2n - 1)(2n + 1 - (2n - 1)) = 8n\)	M1M0
Alt 2 Only \(4n \times 2\)	M0M0
Response only referring to odds and evens or only involving substitution	M0M0A0
Assuming the expression simplifies to \(2k\) and working back could score up to M1M1
Setting up an equation eg \((2n + 1)^2 - (2n - 1)^2 = 4\) could score up to M1M1
For A1 do not allow incorrect use of = eg \(4n^2 + 2n + 2n + 1 - 4n^2 + 2n + 2n - 1\) \(= \dfrac{8n}{4} = 2n\)	M1M1 A0

Specimen 2020 Paper 2 Q21

AQACurrent spec4 marksAlgebraic Proof Expanding Brackets

21 Show that \(\quad (2n + 3)^3 + n^3 \quad\) is divisible by 9 for all integer values of \(n\). [4 marks]

Mark scheme

Answer	Mark	Comments
\(4n^2 + 6n + 6n + 9\) or \(4n^2 + 12n + 9\)	M1	allow one error implied by \(4n^2 + 12n + k\) or \(an^2 + 12n + 9\)
\(8n^3 + 12n^2 + 24n^2 + 36n + 18n + 27\)	M1dep	oe ft their \(4n^2 + 6n + 6n + 9\) allow one error
\(8n^3 + 36n^2 + 54n + 27\) or \(9n^3 + 36n^2 + 54n + 27\)	A1
\(9n^3 + 36n^2 + 54n + 27\) and \(9(n^3 + 4n^2 + 6n + 3)\)	A1	oe eg \((9n^3 + 36n^2 + 54n + 27) \div 9\) \(= n^3 + 4n^2 + 6n + 3\) or \(9n^3 + 36n^2 + 54n + 27\) and all coefficients are divisible by 9

June 2019 Paper 1 Q13

AQAOld spec4 marksAlgebraic Proof Completing the Square

13 Prove that \(\quad (3x + 5)^2 - 5x(x + 10) \geqslant 0 \quad\) for all values of \(x\). [4 marks]

Mark scheme

Answer	Mark	Comments
Alternative method 1
\(9x^2 + 15x + 15x + 25 - 5x^2 - 50x\) or \(9x^2 + 30x + 25 - 5x^2 - 50x\) or \(9x^2 + 15x + 15x + 25\) and \(-5x^2 - 50x\) or \(5x^2 + 50x\)	M1	allow only one error in sign, omission or coefficient but not in more than one of these could be written as 2 separate expansions or in a grid
\(4x^2 - 20x + 25\)	A1
\(4x^2 - 20x + 25\) and \((2x - 5)^2\) or \((2x - 5)(2x - 5)\) or \(4(x - 2.5)^2\) or \(x = 2.5\) or \(b^2 - 4ac = 0\) from quadratic formula	M1dep	factorises or completes the square or uses the quadratic formula correctly. Answer required for M1 dep
\((2x - 5)^2\) or \(4(x - 2.5)^2\) (are squared terms) and so are always \(\geqslant 0\)	A1	oe there must be a stated conclusion eg equal roots and positive quadratic so must be greater than or equal to zero
Alternative method 2
\(9x^2 + 15x + 15x + 25 - 5x^2 - 50x\) or \(9x^2 + 30x + 25 - 5x^2 - 50x\) or \(9x^2 + 15x + 15x + 25\) and \(-5x^2 - 50x\) or \(5x^2 + 50x\)	M1	allow only one error in sign, omission or coefficient but not in more than one of these could be written as 2 separate expansions or in a grid
\(4x^2 - 20x + 25\)	A1
\(4x^2 - 20x + 25\) and \(\dfrac{\mathrm{d}}{\mathrm{d}x} = 8x - 20\) and is zero when \(x = 2.5\)	M1dep	uses calculus to find stationary point
Tests for minimum by using eg \(x = 2\) and \(x = 3\) or by using 2nd derivative or concludes argument by saying this is a positive quadratic curve with minimum point \((2.5,\ 0)\), hence always \(\geqslant 0\)	A1	oe there must be a stated conclusion

June 2018 Paper 2 Q16

AQAOld spec4 marksAlgebraic Proof Expanding Brackets

16 \(A = 2 - 5x \qquad B = 3x - 1 \qquad C = x^2\)

Show that \(\qquad (2A + 3B)^2 \equiv A + B + C\) [4 marks]

Mark scheme

Answer	Mark	Comments
Alternative method 1
\(2(2 - 5x) + 3(3x - 1)\) or \(4 - 10x\) or \(9x - 3\)	M1
\(4 - 10x + 9x - 3 = 1 - x\)	M1dep
\((1 - x)^2 = 1 - 2x + x^2\)	A1	must see working for M2
\(2 - 5x + 3x - 1 + x^2 = 1 - 2x + x^2\)	B1
Alternative method 2
\(4(2 - 5x)^2 + 6(2 - 5x)(3x - 1)\) \(+\ 6(2 - 5x)(3x - 1) + 9(3x - 1)^2\)	M1	oe allow \(+\ 12(2 - 5x)(3x - 1)\) for \(+\ 6(2 - 5x)(3x - 1) + 6(2 - 5x)(3x - 1)\)
\(4(4 - 10x - 10x + 25x^2)\) \(+\ 6(6x - 2 - 15x^2 + 5x)\) \(+\ 6(6x - 2 - 15x^2 + 5x)\) \(+\ 9(9x^2 - 3x - 3x + 1)\) \(= 16 - 40x - 40x + 100x^2 + 36x - 12\) \(-\ 90x^2 + 30x + 36x - 12 - 90x^2\) \(+\ 30x + 81x^2 - 27x - 27x + 9\)	M1dep	oe must see expansions must see working for 1st M1 allow \(+\ 12(6x - 2 - 15x^2 + 5x)\) for \(+\ 6(6x - 2 - 15x^2 + 5x)\) \(+\ 6(6x - 2 - 15x^2 + 5x)\)
\(1 - 2x + x^2\)	A1	must see working for M2
\(2 - 5x + 3x - 1 + x^2 = 1 - 2x + x^2\)	B1
Alternative method 3
\(2(2 - 5x) + 3(3x - 1)\) or \(4 - 10x\) or \(9x - 3\)	M1	oe
\((4 - 10x + 9x - 3)^2\) \(= 16 - 40x + 36x - 12 - 40x + 100x^2\) \(-\ 90x^2 + 30x + 36x - 90x^2 + 81x^2\) \(-\ 27x - 12 + 30x - 27x + 9\)	M1dep	oe must see expansions
\(1 - 2x + x^2\)	A1	must see working for M2
\(2 - 5x + 3x - 1 + x^2 = 1 - 2x + x^2\)	B1

Additional guidance

Allow working down both sides of an equation/identity
M2A1 is for working on \((2A + 3B)^2\)
B1 is for working on \(A + B + C\)
\(1 - 2x + x^2\) with working for M2 seen and \(2 - 5x + 3x - 1 + x^2 = x^2 - 2x + 1\)	4 marks
\(1 - x^2 = 1 - 2x + x^2\) (do not allow missing brackets even if recovered)

June 2016 Paper 2 Q9

AQAOld spec4 marksAlgebraic Proof

9 \(PQ\) is parallel to \(BC\).

PQ and BC are parallel lines. Angle QPB is x. At B the reflex angle PBC is split by line BA into angles y + 100° and 2y + 80°

Not drawn accurately

Prove that \(\quad x = 3y\) [4 marks]

Mark scheme

Answer	Mark	Comments
Alternative method 1 \(PBC\) in terms of \(x\) and in terms of \(y\)
\(180 - x\) or \(360 - (y + 100 + 2y + 80)\) or \(360 - (3y + 180)\)	M1	May be on diagram \(PBC\) (allow \(B\))
\(180 - x\) and \(360 - (y + 100 + 2y + 80)\) or \(360 - (3y + 180)\)	M1
\(180 - x\) and \(180 - 3y\) and \(x = 3y\)	A1	Must have seen correct working for M1 M1
Both reasons given	A1	Must have M1 M1 (Co-)interior angles or allied angles (add up to 180°) and angles at a point (add up to 360°)
Alternative method 2 \(PBC\) in terms of \(x\) + reflex \(PBC\) in terms of \(y\) = 360
\(180 - x\) or \(y + 100 + 2y + 80\) or \(3y + 180\)	M1	May be on diagram \(PBC\) (allow \(B\)) or reflex \(PBC\)
\(180 - x + y + 100 + 2y + 80 = 360\) or \(180 - x + 3y + 180 = 360\)	M1	oe unsimplified correct equation
Simplifies to \(x = 3y\)	A1	Must have seen correct working for M1 M1
Both reasons given	A1	Must have M1 M1 (Co-)interior angles or allied angles (add up to 180°) and angles at a point (add up to 360°)
Alternative method 3 \(x = 180 - PBC\) in terms of \(y\)
\(360 - (y + 100 + 2y + 80)\) or \(360 - (3y + 180)\)	M1	May be on diagram \(PBC\) (allow \(B\))
\(x = 180 - (360 - (y + 100 + 2y + 80))\) \(x = 180 - (360 - (3y + 180))\)	M1	oe unsimplified correct equation
Simplifies to \(x = 3y\)	A1	Must have seen correct working for M1 M1
Both reasons given	A1	Must have M1 M1 (Co-)interior angles or allied angles (add up to 180°) and angles at a point (add up to 360°)
Alternative method 4 \(x + PBC = 180\) and reflex \(PBC\) in terms of \(y + PBC = 360\)
\(x + PBC = 180\) or \(y + 100 + 2y + 80 + PBC = 360\) or \(3y + 180 + PBC = 360\)	M1	\(PBC\) (allow \(B\))
\(x + PBC = 180\) and \(y + 100 + 2y + 80 + PBC = 360\) or \(3y + 180 + PBC = 360\)	M1
\(x + PBC = 180\) and \(3y + PBC = 180\) and \(x = 3y\)	A1	Must have seen correct working for M1 M1
Both reasons given	A1	Must have M1 M1 (Co-)interior angles or allied angles (add up to 180°) and angles at a point (add up to 360°)
Alternative method 5 (Produces \(CB\) to \(X\)) \(PBX\) = reflex \(PBC\) in terms of \(y - 180\)
\(y + 100 + 2y + 80\) or \(3y + 180\)	M1	May be on diagram reflex \(PBC\)
\(y + 100 + 2y + 80 - 180\) or \(3y + 180 - 180\)	M1	\(PBX\)
Simplifies to \(3y\) and states \(x = 3y\)	A1	Must have seen correct working for M1 M1
Both reasons given	A1	Must have M1 M1 Angles on a (straight) line (add up to 180) and alternate angles (are equal)
Alternative method 6 (Produces \(PB\) to \(Y\)) \(CBY\) = reflex \(PBC\) in terms of \(y - 180\)
\(y + 100 + 2y + 80\) or \(3y + 180\)	M1	May be on diagram reflex \(PBC\) (allow reflex \(B\))
\(y + 100 + 2y + 80 - 180\) or \(3y + 180 - 180\)	M1	\(CBY\)
Simplifies to \(3y\) and states \(x = 3y\)	A1	Must have seen correct working for M1 M1
Both reasons given	A1	Must have M1 M1 Angles on a (straight) line (add up to 180) and corresponding angles (are equal)
Alternative method 7 (Produces \(CB\) to \(X\)) \(PBX = 180 - PBC\) in terms of \(y\)
\(360 - (y + 100 + 2y + 80)\) or \(360 - (3y + 180)\)	M1	May be on diagram \(PBC\) (allow \(B\))
\(180 - (360 - (y + 100 + 2y + 80))\) or \(180 - (360 - (3y + 180))\)	M1	\(PBX\)
Simplifies to \(3y\) and states \(x = 3y\)	A1	Must have seen correct working for M1 M1
All reasons given	A1	Must have M1 M1 Angles at a point and angles on a (straight) line (add up to 180) and alternate angles (are equal)
Alternative method 8 (Produces \(PB\) to \(Y\)) \(CBY = 180 - PBC\) in terms of \(y\)
\(360 - (y + 100 + 2y + 80)\) or \(360 - (3y + 180)\)	M1	May be on diagram \(PBC\) (allow \(B\))
\(180 - (360 - (y + 100 + 2y + 80))\) or \(180 - (360 - (3y + 180))\)	M1	\(CBY\)
Simplifies to \(3y\) and states \(x = 3y\)	A1	Must have seen correct working for M1 M1
All reasons given	A1	Must have M1 M1 Angles at a point and angles on a (straight) line (add up to 180) and corresponding angles (are equal)

Additional guidance

Recovery of brackets is not allowed as it is a proof

Acceptable reasons must include the word ‘angles’
Angles at a point can be angles round a point

These reasons are not allowed:
Alternating angles, alternative angles, angles in a circle, straight line,
at a point, round a point, parallel lines

Supplementary angles are not allowed unless accompanied by an acceptable reason eg Allied angles are supplementary is allowed

Other variations on these methods will be seen. Escalate if necessary.

Starting with \(x = 3y\) or substituting values for \(x\) and \(y\) is zero unless M marks seen in working

Algebraic Proof

AQA

AQA · Old spec

Additional guidance

Additional guidance

Additional guidance

Additional guidance

Additional guidance