AQA Further Paper 2 2019 June — Question 10 7 marks

Exam BoardAQA
ModuleFurther Paper 2 (Further Paper 2)
Year2019
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicProof by induction
TypeProve divisibility
DifficultyStandard +0.3 This is a standard proof by induction with a divisibility statement. While it requires proper induction structure (base case, assumption, inductive step), the algebra is straightforward: f(k+1) - f(k) = 3kΒ² + 9k + 11 needs checking, and showing 6|f(k+1) given 6|f(k) involves routine factoring. It's slightly above average difficulty due to being a proof question requiring formal structure, but it's a textbook-style induction problem without novel insight or tricky algebraic manipulation.
Spec4.01a Mathematical induction: construct proofs
Prove by induction that \(f(n) = n^3 + 3n^2 + 8n\) is divisible by 6 for all integers \(n \geq 1\) [7 marks]
Question 10:
AnswerMarks
10Demonstrates the result for
and states that it is
AnswerMarks Guidance
true forAO1.1b B1
so the result is true for
Assu𝑛𝑛m=e1 the resf(u1lt) is= tr1u2e= fo2r Γ—6 :
Then for som 𝑛𝑛e= in1teger m
𝑛𝑛 = π‘˜π‘˜
f(π‘˜π‘˜)= 6π‘šπ‘š 3 2
f(π‘˜π‘˜+1)=(π‘˜π‘˜+1) +3(π‘˜π‘˜+1) +8(π‘˜π‘˜+1)
3 2 2
=π‘˜π‘˜ +3π‘˜π‘˜ +3π‘˜π‘˜+1+3(π‘˜π‘˜ +2π‘˜π‘˜+1)+8(π‘˜π‘˜+1)
3 2
3 = π‘˜π‘˜ 2 +6π‘˜π‘˜ +17π‘˜π‘˜+1 3 2 2
f(π‘˜π‘˜+1)=π‘˜π‘˜ +6π‘˜π‘˜ +17π‘˜π‘˜ +12βˆ’(π‘˜π‘˜ +3π‘˜π‘˜ +8π‘˜π‘˜)
2
= i6sπ‘šπ‘š a +m3uπ‘˜π‘˜ltip+le9 π‘˜π‘˜of+ 31 2
2
3π‘˜π‘˜ +9π‘˜π‘˜+12
2 a2nd one of or is
e3vπ‘˜π‘˜ 2en+ 9π‘˜π‘˜+12=3(π‘˜π‘˜ +3π‘˜π‘˜)+12
π‘˜π‘˜ +3π‘˜π‘˜ = π‘˜π‘˜is( π‘˜π‘˜ev+en3 )and π‘˜π‘˜ π‘˜π‘˜+is3 an
even2 multiple of 3 and henc2e divisible by 6
βˆ΄π‘˜π‘˜ +3π‘˜π‘˜ 3(π‘˜π‘˜ +3π‘˜π‘˜+4)
is divisible by 6 if is
divisible by 6
∴ f(π‘˜π‘˜+1) f(π‘˜π‘˜)
We also know that is divisible by 6, so by
induction this completes the proof.
f(1)
𝑛𝑛As=su1mes the result true for
AnswerMarks Guidance
𝑛𝑛 = 1AO2.4 M1
Obtains the difference
𝑛𝑛 = π‘˜π‘˜
AnswerMarks Guidance
be tw een andAO3.1a M1
Calculates the difference
between f(π‘˜π‘˜+1) f(π‘˜π‘˜)
and correctly
AnswerMarks Guidance
f(π‘˜π‘˜+1)AO1.1b A1
Ded uf(cπ‘˜π‘˜e)s that the difference
AnswerMarks Guidance
is a multiple of 3AO2.2a M1
Deduces that the difference
AnswerMarks Guidance
is a multiple of 2AO2.2a M1
Completes a rigorous
argument and explains how
their argument proves the
AnswerMarks Guidance
required resultAO2.1 R1
Total7
QMarking Instructions AO 
Question 10:
10 | Demonstrates the result for
and states that it is
true for | AO1.1b | B1 | Let then
so the result is true for
Assu𝑛𝑛m=e1 the resf(u1lt) is= tr1u2e= fo2r Γ—6 :
Then for som 𝑛𝑛e= in1teger m
𝑛𝑛 = π‘˜π‘˜
f(π‘˜π‘˜)= 6π‘šπ‘š 3 2
f(π‘˜π‘˜+1)=(π‘˜π‘˜+1) +3(π‘˜π‘˜+1) +8(π‘˜π‘˜+1)
3 2 2
=π‘˜π‘˜ +3π‘˜π‘˜ +3π‘˜π‘˜+1+3(π‘˜π‘˜ +2π‘˜π‘˜+1)+8(π‘˜π‘˜+1)
3 2
3 = π‘˜π‘˜ 2 +6π‘˜π‘˜ +17π‘˜π‘˜+1 3 2 2
f(π‘˜π‘˜+1)=π‘˜π‘˜ +6π‘˜π‘˜ +17π‘˜π‘˜ +12βˆ’(π‘˜π‘˜ +3π‘˜π‘˜ +8π‘˜π‘˜)
2
= i6sπ‘šπ‘š a +m3uπ‘˜π‘˜ltip+le9 π‘˜π‘˜of+ 31 2
2
3π‘˜π‘˜ +9π‘˜π‘˜+12
2 a2nd one of or is
e3vπ‘˜π‘˜ 2en+ 9π‘˜π‘˜+12=3(π‘˜π‘˜ +3π‘˜π‘˜)+12
π‘˜π‘˜ +3π‘˜π‘˜ = π‘˜π‘˜is( π‘˜π‘˜ev+en3 )and π‘˜π‘˜ π‘˜π‘˜+is3 an
even2 multiple of 3 and henc2e divisible by 6
βˆ΄π‘˜π‘˜ +3π‘˜π‘˜ 3(π‘˜π‘˜ +3π‘˜π‘˜+4)
is divisible by 6 if is
divisible by 6
∴ f(π‘˜π‘˜+1) f(π‘˜π‘˜)
We also know that is divisible by 6, so by
induction this completes the proof.
f(1)
𝑛𝑛As=su1mes the result true for
𝑛𝑛 = 1 | AO2.4 | M1
Obtains the difference
𝑛𝑛 = π‘˜π‘˜
be tw een and | AO3.1a | M1
Calculates the difference
between f(π‘˜π‘˜+1) f(π‘˜π‘˜)
and correctly
f(π‘˜π‘˜+1) | AO1.1b | A1
Ded uf(cπ‘˜π‘˜e)s that the difference
is a multiple of 3 | AO2.2a | M1
Deduces that the difference
is a multiple of 2 | AO2.2a | M1
Completes a rigorous
argument and explains how
their argument proves the
required result | AO2.1 | R1
Total | 7
Q | Marking Instructions | AO | Marks | Typical solution
Prove by induction that $f(n) = n^3 + 3n^2 + 8n$ is divisible by 6 for all integers $n \geq 1$
[7 marks]

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