AQA Further Paper 2 2020 June — Question 10 6 marks

Exam BoardAQA
ModuleFurther Paper 2 (Further Paper 2)
Year2020
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicProof by induction
TypeProve recurrence relation formula
DifficultyChallenging +1.2 This is a standard proof by induction on a recurrence relation with 6 marks. While it requires careful algebraic manipulation to verify the inductive step (substituting the formula into the recurrence relation and simplifying to show it holds for n+1), the structure is entirely routine for Further Maths students. The algebra involves fraction manipulation but no particularly deep insightβ€”it's a textbook-style induction proof that's more demanding than basic C1/C2 questions but well within expected Further Maths territory.
Spec1.04e Sequences: nth term and recurrence relations4.01a Mathematical induction: construct proofs
The sequence \(u_1, u_2, u_3, \ldots\) is defined by $$u_1 = 0 \quad u_{n+1} = \frac{5}{6 - u_n}$$ Prove by induction that, for all integers \(n \geq 1\), $$u_n = \frac{5^n - 5}{5^n - 1}$$ [6 marks]
Question 10:
AnswerMarks
10Shows that is
𝑛𝑛
5 βˆ’5
true for
AnswerMarks Guidance
𝑒𝑒𝑛𝑛 = 5 𝑛𝑛 βˆ’11.1b B1
𝑛𝑛 = 1
1
5 βˆ’5
𝑒𝑒1 = 5 1 βˆ’1 = 0
so the result is true for
Assume the result is tru 𝑛𝑛e= fo1r :
𝑛𝑛 = π‘˜π‘˜
Then
5
π‘’π‘’π‘˜π‘˜+1 = 5π‘˜π‘˜βˆ’5
6βˆ’οΏ½5π‘˜π‘˜βˆ’1οΏ½
π‘˜π‘˜ π‘˜π‘˜ π‘˜π‘˜
5 βˆ’5 6οΏ½5 βˆ’1οΏ½βˆ’οΏ½5 βˆ’5οΏ½
6βˆ’οΏ½ π‘˜π‘˜ οΏ½ = π‘˜π‘˜
5 βˆ’1 5 βˆ’1
π‘˜π‘˜ π‘˜π‘˜
6Γ—5 βˆ’5 βˆ’6+5
= π‘˜π‘˜
5 βˆ’1
π‘˜π‘˜ π‘˜π‘˜+1
5Γ—5 βˆ’1 5 βˆ’1
= π‘˜π‘˜ = π‘˜π‘˜
5 βˆ’1 5 βˆ’1
π‘˜π‘˜ π‘˜π‘˜+1
5 βˆ’1 5 βˆ’5
∴ π‘’π‘’π‘˜π‘˜+1 = 5Γ— π‘˜π‘˜+1 = π‘˜π‘˜+1
and the result also 5holdsβˆ’ fo1r 5 βˆ’1
The formula for is true for𝑛𝑛 = π‘˜π‘˜+ if1 true
for , then it’s also true for
𝑒𝑒𝑛𝑛 𝑛𝑛 = 1;
and hence by induction for
𝑛𝑛 = π‘˜π‘˜ 𝑛𝑛𝑛𝑛 = π‘˜π‘˜+1
States th𝑛𝑛e= as1sumption that
is true for
𝑛𝑛
AnswerMarks Guidance
5 βˆ’52.4 M1
𝑒𝑒𝑛𝑛 = 5 𝑛𝑛 βˆ’1 𝑛𝑛 = π‘˜π‘˜
Uses the recurrence
relation and the
assumption to express
AnswerMarks Guidance
in terms of3.1a M1
𝑒𝑒E π‘˜π‘˜ x + p 1 resses π‘˜π‘˜as a
single fraction.
AnswerMarks Guidance
π‘’π‘’π‘˜π‘˜+11.1a M1
Completes rigorous
working to deduce that
π‘˜π‘˜+1
AnswerMarks Guidance
5 βˆ’52.2a R1
π‘’π‘’π‘˜π‘˜+1 = π‘˜π‘˜+1
Concludes a 5reasoβˆ’ne1d
argument by stating that
the formula for is true
for that if true for
, then it’s 𝑒𝑒a 𝑛𝑛 lso true
for 𝑛𝑛 = 1; and hence
𝑛𝑛 = π‘˜π‘˜
by induction for
𝑛𝑛 = π‘˜π‘˜+1 𝑛𝑛
5 βˆ’5
AnswerMarks Guidance
𝑒𝑒𝑛𝑛 = 5 𝑛𝑛 βˆ’12.1 R1

Total

AnswerMarks Guidance
𝑛𝑛 β‰₯ 16 5 βˆ’5
𝑛𝑛 𝑛𝑛
5 βˆ’1
AnswerMarks Guidance
QMarking Instructions AO 
Question 10:
10 | Shows that is
𝑛𝑛
5 βˆ’5
true for
𝑒𝑒𝑛𝑛 = 5 𝑛𝑛 βˆ’1 | 1.1b | B1 | Let ; then the formula gives
𝑛𝑛 = 1
1
5 βˆ’5
𝑒𝑒1 = 5 1 βˆ’1 = 0
so the result is true for
Assume the result is tru 𝑛𝑛e= fo1r :
𝑛𝑛 = π‘˜π‘˜
Then
5
π‘’π‘’π‘˜π‘˜+1 = 5π‘˜π‘˜βˆ’5
6βˆ’οΏ½5π‘˜π‘˜βˆ’1οΏ½
π‘˜π‘˜ π‘˜π‘˜ π‘˜π‘˜
5 βˆ’5 6οΏ½5 βˆ’1οΏ½βˆ’οΏ½5 βˆ’5οΏ½
6βˆ’οΏ½ π‘˜π‘˜ οΏ½ = π‘˜π‘˜
5 βˆ’1 5 βˆ’1
π‘˜π‘˜ π‘˜π‘˜
6Γ—5 βˆ’5 βˆ’6+5
= π‘˜π‘˜
5 βˆ’1
π‘˜π‘˜ π‘˜π‘˜+1
5Γ—5 βˆ’1 5 βˆ’1
= π‘˜π‘˜ = π‘˜π‘˜
5 βˆ’1 5 βˆ’1
π‘˜π‘˜ π‘˜π‘˜+1
5 βˆ’1 5 βˆ’5
∴ π‘’π‘’π‘˜π‘˜+1 = 5Γ— π‘˜π‘˜+1 = π‘˜π‘˜+1
and the result also 5holdsβˆ’ fo1r 5 βˆ’1
The formula for is true for𝑛𝑛 = π‘˜π‘˜+ if1 true
for , then it’s also true for
𝑒𝑒𝑛𝑛 𝑛𝑛 = 1;
and hence by induction for
𝑛𝑛 = π‘˜π‘˜ 𝑛𝑛𝑛𝑛 = π‘˜π‘˜+1
States th𝑛𝑛e= as1sumption that
is true for
𝑛𝑛
5 βˆ’5 | 2.4 | M1
𝑒𝑒𝑛𝑛 = 5 𝑛𝑛 βˆ’1 𝑛𝑛 = π‘˜π‘˜
Uses the recurrence
relation and the
assumption to express
in terms of | 3.1a | M1
𝑒𝑒E π‘˜π‘˜ x + p 1 resses π‘˜π‘˜as a
single fraction.
π‘’π‘’π‘˜π‘˜+1 | 1.1a | M1
Completes rigorous
working to deduce that
π‘˜π‘˜+1
5 βˆ’5 | 2.2a | R1
π‘’π‘’π‘˜π‘˜+1 = π‘˜π‘˜+1
Concludes a 5reasoβˆ’ne1d
argument by stating that
the formula for is true
for that if true for
, then it’s 𝑒𝑒a 𝑛𝑛 lso true
for 𝑛𝑛 = 1; and hence
𝑛𝑛 = π‘˜π‘˜
by induction for
𝑛𝑛 = π‘˜π‘˜+1 𝑛𝑛
5 βˆ’5
𝑒𝑒𝑛𝑛 = 5 𝑛𝑛 βˆ’1 | 2.1 | R1
Total
𝑛𝑛 β‰₯ 1 | 6 | 5 βˆ’5
𝑛𝑛 𝑛𝑛
5 βˆ’1
Q | Marking Instructions | AO | Marks | Typical Solution
The sequence $u_1, u_2, u_3, \ldots$ is defined by
$$u_1 = 0 \quad u_{n+1} = \frac{5}{6 - u_n}$$

Prove by induction that, for all integers $n \geq 1$,
$$u_n = \frac{5^n - 5}{5^n - 1}$$
[6 marks]

\hfill \mbox{\textit{AQA Further Paper 2 2020 Q10 [6]}}
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