AQA Further AS Paper 1 2020 June — Question 7 4 marks

Exam BoardAQA
ModuleFurther AS Paper 1 (Further AS Paper 1)
Year2020
SessionJune
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicProof by induction
TypeProve divisibility
DifficultyModerate -0.3 This is a straightforward proof by induction with a simple divisibility claim. The base case is trivial (7-3=4), and the inductive step requires only basic algebraic manipulation (7^{k+1}-3^{k+1} = 7Β·7^k - 3Β·3^k = 7(7^k-3^k) + 4Β·7^k). While it's a Further Maths question, divisibility induction proofs are standard exercises that follow a well-rehearsed template with no novel insight required.
Spec4.01a Mathematical induction: construct proofs
Prove by induction that, for all integers \(n \geq 1\), the expression \(7^n - 3^n\) is divisible by 4 [4 marks]
Question 7:
AnswerMarks Guidance
7Shows that is divisible by 4 for .
𝑛𝑛 𝑛𝑛1.1b B1
Ass7umβˆ’e 3it is= tr7uβˆ’e f3or= 4
𝑛𝑛 = π‘˜π‘˜
where π‘˜π‘˜ is aπ‘˜π‘˜n integer
∴ 7 βˆ’3 = 4π‘šπ‘š
π‘šπ‘š
π‘˜π‘˜+1 π‘˜π‘˜+1 π‘˜π‘˜ π‘˜π‘˜
7 βˆ’3 = 7Γ—7 βˆ’3Γ—3
π‘˜π‘˜ π‘˜π‘˜
= 7οΏ½4π‘šπ‘š+3 οΏ½βˆ’3Γ—3
π‘˜π‘˜
= 28π‘šπ‘š+4Γ—3
π‘˜π‘˜
it is also tr = ue 4 οΏ½fo 7 r π‘šπ‘š +3 οΏ½
It is true for ∴ . If it is true for 𝑛𝑛 = π‘˜π‘˜ +the1n it is true for
. Therefore, by induction, is divisible
by 4 for all in𝑛𝑛te=ge1rs, . 𝑛𝑛 = 𝑛𝑛 π‘˜π‘˜ 𝑛𝑛
𝑛𝑛 = π‘˜π‘˜+1 7 βˆ’3
𝑛𝑛 β‰₯ 1
7 βˆ’3 𝑛𝑛 = 1
States the assumption that
is divisible by 4 and considers ,
π‘˜π‘˜ π‘˜π‘˜ π‘˜π‘˜+1 π‘˜π‘˜+1
b7y uβˆ’si3ng or . 7 βˆ’3
AnswerMarks Guidance
π‘˜π‘˜ π‘˜π‘˜2.4 M1
7Γ—7 3Γ—3
Completes rigorous working to deduce that is divisible by 4.
AnswerMarks Guidance
π‘˜π‘˜+1 π‘˜π‘˜+12.2a R1
7 βˆ’3
Concludes a reasoned argument by stating that
is divisible by 4 for ;
𝑛𝑛 𝑛𝑛
t7hatβˆ’ if 3 is divisible by 𝑛𝑛4,= th1en is divisible by 4
π‘˜π‘˜ π‘˜π‘˜ π‘˜π‘˜+1 π‘˜π‘˜+1
and he7ncβˆ’e, 3by induction, is d7ivisibβˆ’le3 by 4 for .
AnswerMarks Guidance
𝑛𝑛 𝑛𝑛2.1 R1
7 βˆ’3 𝑛𝑛 β‰₯ 1
AnswerMarks Guidance
Total4
QMarking instructions AO 
Question 7:
7 | Shows that is divisible by 4 for .
𝑛𝑛 𝑛𝑛 | 1.1b | B1 | 1 1
Ass7umβˆ’e 3it is= tr7uβˆ’e f3or= 4
𝑛𝑛 = π‘˜π‘˜
where π‘˜π‘˜ is aπ‘˜π‘˜n integer
∴ 7 βˆ’3 = 4π‘šπ‘š
π‘šπ‘š
π‘˜π‘˜+1 π‘˜π‘˜+1 π‘˜π‘˜ π‘˜π‘˜
7 βˆ’3 = 7Γ—7 βˆ’3Γ—3
π‘˜π‘˜ π‘˜π‘˜
= 7οΏ½4π‘šπ‘š+3 οΏ½βˆ’3Γ—3
π‘˜π‘˜
= 28π‘šπ‘š+4Γ—3
π‘˜π‘˜
it is also tr = ue 4 οΏ½fo 7 r π‘šπ‘š +3 οΏ½
It is true for ∴ . If it is true for 𝑛𝑛 = π‘˜π‘˜ +the1n it is true for
. Therefore, by induction, is divisible
by 4 for all in𝑛𝑛te=ge1rs, . 𝑛𝑛 = 𝑛𝑛 π‘˜π‘˜ 𝑛𝑛
𝑛𝑛 = π‘˜π‘˜+1 7 βˆ’3
𝑛𝑛 β‰₯ 1
7 βˆ’3 𝑛𝑛 = 1
States the assumption that
is divisible by 4 and considers ,
π‘˜π‘˜ π‘˜π‘˜ π‘˜π‘˜+1 π‘˜π‘˜+1
b7y uβˆ’si3ng or . 7 βˆ’3
π‘˜π‘˜ π‘˜π‘˜ | 2.4 | M1
7Γ—7 3Γ—3
Completes rigorous working to deduce that is divisible by 4.
π‘˜π‘˜+1 π‘˜π‘˜+1 | 2.2a | R1
7 βˆ’3
Concludes a reasoned argument by stating that
is divisible by 4 for ;
𝑛𝑛 𝑛𝑛
t7hatβˆ’ if 3 is divisible by 𝑛𝑛4,= th1en is divisible by 4
π‘˜π‘˜ π‘˜π‘˜ π‘˜π‘˜+1 π‘˜π‘˜+1
and he7ncβˆ’e, 3by induction, is d7ivisibβˆ’le3 by 4 for .
𝑛𝑛 𝑛𝑛 | 2.1 | R1
7 βˆ’3 𝑛𝑛 β‰₯ 1
Total | 4
Q | Marking instructions | AO | Marks | Typical solution
Prove by induction that, for all integers $n \geq 1$, the expression $7^n - 3^n$ is divisible by 4
[4 marks]

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