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General algebraic proofs

3 exam-style questions. Answers and the mark for each step are on the last page.

Question 1Non-calculator · 3 marks
$n$ is an integer.
(a) Prove that the sum of any three consecutive even numbers is a multiple of 6. (3)
Question 2Non-calculator · 4 marks
A pupil claims that the sum of any three consecutive integers is divisible by 6.
(a) Prove that the sum is always divisible by 3. (3)
(b) Use a counterexample to disprove the pupil’s claim about 6. (1)
Question 3Calculator · 3 marks
(a) Prove that $n^{4}$ $-$ 1 is divisible by 16 whenever n is an odd integer. (3)

Answers and marks

Question 1

(a) $2n + (2n + 2) + (2n + 4) = 6n + 6 = 6(n + 1)$

M1 Writing consecutive even numbers as $2n$ , $2n + 2$ , $2n + 4$ .
M1 Simplifying to $6n + 6$ .
C1 Writing $6(n + 1)$ and concluding that it is a multiple of 6.

Question 2

(a) Writing the integers as n-1, n and n+1 gives sum 3n, a multiple of 3.

M1 Represent the consecutive integers around their middle integer.
M1 Collect terms into an integer multiple.
C1 Correct conclusion with supporting reasoning: Writing the integers as n-1, n and n+1 gives sum 3n, a multiple of 3.

(b) 2+3+4=9, which is not divisible by 6.

C1 Correct conclusion with supporting reasoning: 2+3+4=9, which is not divisible by 6.

Question 3

(a) For n = 2k + 1, $n^{2}$ $-$ 1 = 4k(k + 1) is a multiple of 8. Also $n^{2}$ + 1 is even. Hence $n^{4}$ $-$ 1 = ( $n^{2}$ $-$ 1)( $n^{2}$ + 1) is a multiple of 16.

P1 Establishing $(2k+1)^{2}-1=4k(k+1)$ or an equivalent valid method.
P1 Establishing $n^{4}-1=(n^{2}-1)(n^{2}+1)$ or an equivalent valid method.
C1 Correct conclusion with the complete supporting argument: For n = 2k + 1, $n^{2}$ $-$ 1 = 4k(k + 1) is a multiple of 8. Also $n^{2}$ + 1 is even. Hence $n^{4}$ $-$ 1 = ( $n^{2}$ $-$ 1)( $n^{2}$ + 1) is a multiple of 16.

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General algebraic proofs

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