AQA Further Paper 2 Specimen — Question 6 5 marks

Exam BoardAQA
ModuleFurther Paper 2 (Further Paper 2)
SessionSpecimen
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicProof by induction
TypeProve divisibility
DifficultyStandard +0.3 This is a standard proof by induction question with straightforward algebra. While it requires knowledge of the induction framework and some manipulation of powers, the algebraic steps are routine (factoring out 7 from 8^{k+1} - 8^k) and the question follows a well-practiced template. It's slightly above average difficulty only because it's a proof question requiring formal structure, but it's easier than most Further Maths content.
Spec1.01a Proof: structure of mathematical proof and logical steps4.01a Mathematical induction: construct proofs
Prove that \(8^n - 7n + 6\) is divisible by 7 for all integers \(n \geq 0\) [5 marks]
Question 6:
AnswerMarks
6Uses proof by induction and
investigates the expression
for n = 0 and n = k (must
see evidence of both n = 0
AnswerMarks Guidance
and n = k being considered)AO3.1a B1
f(0)167
f(n) is divisible by 7 when n = 0
Consider n = k
Assume that f(k) is divisible by 7
f(k1)8k17(k1)6
fk18fk56k7k1648
fk18fk49k49
fk 18fk49k 1
8fk7(7k 7)
f(k 1) is divisible by 7 since f(k)
is divisible by 7
Therefore
f(k) is divisible by 7  f(k + 1) is divisible
by 7
Since f(0) is divisible by 7 and
f(k) is divisible by 7 ⇒f(k + 1) is divisible
by 7
then, by induction, f(n)8n 7n6 is
divisible by 7 for all integers n 0
Shows that statement is true
AnswerMarks Guidance
for n = 0AO1.1b B1
Commences argument by
considering f(k + 1) in
AnswerMarks Guidance
terms of f(k)AO2.1 R1
Makes correct deduction
that if f(n) is divisible by 7
then f(n + 1) is also divisible
AnswerMarks Guidance
by 7AO2.2a R1
Completes a rigorous
argument and explains how
their argument proves the
AnswerMarks Guidance
required result. AGAO2.4 R1
Total5
QMarking Instructions AO
6Let f(n)8n7n6
ALT
f(0)167
f(n) is divisible by 7 when n=0
Consider n = k
Assume that f(k) is divisible by 7
f(k1)8k17(k1)6
8  8k 7k6  87k7k148
8fk49k49
8fk7(7k7)
f(k 1) is divisible by 7 since f(k)
is divisible by 7
Therefore
f(k) is divisible by 7 f(k + 1) is divisible
by 7
Since f(0) is divisible by 7 and
f(k) is divisible by 7 ⇒f(k + 1) is divisible
by 7
then, by induction, f(n)8n7n6 is
divisible by 7 for all integers n0
AnswerMarks Guidance
Total5
QMarking Instructions AO 
Question 6:
6 | Uses proof by induction and
investigates the expression
for n = 0 and n = k (must
see evidence of both n = 0
and n = k being considered) | AO3.1a | B1 | Let f(n)8n 7n6
f(0)167
f(n) is divisible by 7 when n = 0
Consider n = k
Assume that f(k) is divisible by 7
f(k1)8k17(k1)6
fk18fk56k7k1648
fk18fk49k49
fk 18fk49k 1
8fk7(7k 7)
f(k 1) is divisible by 7 since f(k)
is divisible by 7
Therefore
f(k) is divisible by 7  f(k + 1) is divisible
by 7
Since f(0) is divisible by 7 and
f(k) is divisible by 7 ⇒f(k + 1) is divisible
by 7
then, by induction, f(n)8n 7n6 is
divisible by 7 for all integers n 0
Shows that statement is true
for n = 0 | AO1.1b | B1
Commences argument by
considering f(k + 1) in
terms of f(k) | AO2.1 | R1
Makes correct deduction
that if f(n) is divisible by 7
then f(n + 1) is also divisible
by 7 | AO2.2a | R1
Completes a rigorous
argument and explains how
their argument proves the
required result. AG | AO2.4 | R1
Total | 5
Q | Marking Instructions | AO | Marks | Typical Solution
6 | Let f(n)8n7n6
ALT
f(0)167
f(n) is divisible by 7 when n=0
Consider n = k
Assume that f(k) is divisible by 7
f(k1)8k17(k1)6
8  8k 7k6  87k7k148
8fk49k49
8fk7(7k7)
f(k 1) is divisible by 7 since f(k)
is divisible by 7
Therefore
f(k) is divisible by 7 f(k + 1) is divisible
by 7
Since f(0) is divisible by 7 and
f(k) is divisible by 7 ⇒f(k + 1) is divisible
by 7
then, by induction, f(n)8n7n6 is
divisible by 7 for all integers n0
Total | 5
Q | Marking Instructions | AO | Marks | Typical Solution
Prove that $8^n - 7n + 6$ is divisible by 7 for all integers $n \geq 0$
[5 marks]

\hfill \mbox{\textit{AQA Further Paper 2  Q6 [5]}}
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