CAIE Further Paper 1 2021 June — Question 1 6 marks

Exam BoardCAIE
ModuleFurther Paper 1 (Further Paper 1)
Year2021
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicProof by induction
TypeProve divisibility
DifficultyStandard +0.3 This is a standard divisibility proof by induction with straightforward algebra. The inductive step requires factoring out 15 from 16·2^(4n) + 30·31^n, which is routine manipulation. While it's a Further Maths topic, it follows the standard template without requiring novel insight or particularly complex algebraic manipulation.
Spec4.01a Mathematical induction: construct proofs

1 Prove by mathematical induction that \(2 ^ { 4 n } + 31 ^ { n } - 2\) is divisible by 15 for all positive integers \(n\).

Question 1:
AnswerMarks Guidance
AnswerMarks Guidance
\(2^4 + 31 - 2 = 45\) is divisible by 15B1 Checks base case
Assume that \(2^{4k} + 31^k - 2\) is divisible by 15 for some positive integer \(k\)B1 States inductive hypothesis
Then \(2^{4k+4} + 31^{k+1} - 2 = (15+1)2^{4k} + (30+1)31^k - 2\)M1 A1 Separates \(2^{4k} + 31^k - 2\) or considers difference
is divisible by 15 because \(15 \times 2^{4k} + 30 \times 31^k\) is divisible by 15A1
Hence, by induction, true for every positive integer \(n\)A1
Total6
## Question 1:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $2^4 + 31 - 2 = 45$ is divisible by 15 | **B1** | Checks base case |
| Assume that $2^{4k} + 31^k - 2$ is divisible by 15 for some positive integer $k$ | **B1** | States inductive hypothesis |
| Then $2^{4k+4} + 31^{k+1} - 2 = (15+1)2^{4k} + (30+1)31^k - 2$ | **M1 A1** | Separates $2^{4k} + 31^k - 2$ or considers difference |
| is divisible by 15 because $15 \times 2^{4k} + 30 \times 31^k$ is divisible by 15 | **A1** | |
| Hence, by induction, true for every positive integer $n$ | **A1** | |
| **Total** | **6** | |

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1 Prove by mathematical induction that $2 ^ { 4 n } + 31 ^ { n } - 2$ is divisible by 15 for all positive integers $n$.\\

\hfill \mbox{\textit{CAIE Further Paper 1 2021 Q1 [6]}}