CAIE Further Paper 1 2020 Specimen — Question 2 7 marks

Exam BoardCAIE
ModuleFurther Paper 1 (Further Paper 1)
Year2020
SessionSpecimen
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicProof by induction
TypeProve divisibility
DifficultyStandard +0.3 This is a straightforward proof by induction showing 5^n(4n+1) is divisible by a number for all positive integers n. It requires standard induction technique (base case, inductive step with algebraic manipulation) but involves routine factorization and no novel insight—slightly easier than average for Further Maths.
Spec4.01a Mathematical induction: construct proofs
2 It is given that \(\phi ( n ) = 5 ^ { n } ( 4 n + 1 ) - 1\), for \(n = 1,2,3 , \ldots\).
Prove, by mathematical induction, that \(\phi ( n )\) is divisible by 8 for every positive integer \(n\).
Question 2:
AnswerMarks Guidance
AnswerMarks Guidance
\(\phi(1) = 5\times5 - 1 = 24\) which is divisible by 8; \(P_1\) true(B1) EITHER Solution 1
Assume \(P_k\) true for some positive integer \(k\): so \(\phi(k) = 8l\)B1
\(\phi(k+1) - \phi(k)\)M1
\(= 5^{k+1}(4k+5) - 1 - 5^k(4k+1) + 1\)A1
\(= 5^k(16k+24) = 8m\)A1
\(\phi(k+1) = 8(l+m)\)A1
Hence, by PMI, true for all positive integersA1)
\(\phi(1) = 5\times5-1 = 24\) divisible by 8; \(P_1\) true(B1) OR Solution 2
Assume \(P_k\) true for some positive integer \(k\): so \(\phi(k) = 8l\)B1
\(\phi(k+1) = 5^{k+1}(4k+5)-1\)M1A1
\(= 5\times(4k\times5^k + 5^k) - 1\)A1
\(= 40l + 20\times5^k + 4\)A1
\(= 40l + 4(5^{k+1}+1)\)
\(= 40l + 4(2m)\) since \((5^{k+1}+1)\) is even, so is divisible by 8A1 Or equivalent argument
Hence, by PMI, true for all positive integersA1) 
## Question 2:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\phi(1) = 5\times5 - 1 = 24$ which is divisible by 8; $P_1$ true | (B1) | EITHER Solution 1 |
| Assume $P_k$ true for some positive integer $k$: so $\phi(k) = 8l$ | B1 | |
| $\phi(k+1) - \phi(k)$ | M1 | |
| $= 5^{k+1}(4k+5) - 1 - 5^k(4k+1) + 1$ | A1 | |
| $= 5^k(16k+24) = 8m$ | A1 | |
| $\phi(k+1) = 8(l+m)$ | A1 | |
| Hence, by PMI, true for all positive integers | A1) | |
| $\phi(1) = 5\times5-1 = 24$ divisible by 8; $P_1$ true | (B1) | OR Solution 2 |
| Assume $P_k$ true for some positive integer $k$: so $\phi(k) = 8l$ | B1 | |
| $\phi(k+1) = 5^{k+1}(4k+5)-1$ | M1A1 | |
| $= 5\times(4k\times5^k + 5^k) - 1$ | A1 | |
| $= 40l + 20\times5^k + 4$ | A1 | |
| $= 40l + 4(5^{k+1}+1)$ | | |
| $= 40l + 4(2m)$ since $(5^{k+1}+1)$ is even, so is divisible by 8 | A1 | Or equivalent argument |
| Hence, by PMI, true for all positive integers | A1) | |

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2 It is given that $\phi ( n ) = 5 ^ { n } ( 4 n + 1 ) - 1$, for $n = 1,2,3 , \ldots$.\\
Prove, by mathematical induction, that $\phi ( n )$ is divisible by 8 for every positive integer $n$.\\

\hfill \mbox{\textit{CAIE Further Paper 1 2020 Q2 [7]}}
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