CAIE FP1 2016 June — Question 3 6 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2016
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicProof by induction
TypeProve divisibility
DifficultyStandard +0.3 This is a standard proof by induction for divisibility, requiring students to verify the base case (n=1), assume for n=k, and prove for n=k+1 by algebraic manipulation. While it involves Further Maths content, the technique is routine and the algebra is straightforward (factoring out 9 from the difference). Slightly above average difficulty due to being Further Maths and requiring careful algebraic manipulation, but follows a well-practiced template.
Spec4.01a Mathematical induction: construct proofs
3 Prove by mathematical induction that, for all positive integers \(n , 10 ^ { n } + 3 \times 4 ^ { n + 2 } + 5\) is divisible by 9 .
Question 3:
AnswerMarks Guidance
Working/AnswerMarks Guidance
For \(n=1\): \(10 + 192 + 5 = 207 = 9 \times 23 \Rightarrow H_1\) is trueB1
Assume \(H_k\) is true for some positive integer \(k \Rightarrow 10^n + 3 \cdot 4^{n+2} + 5 = 9\alpha\)B1
Hence \(f(n+1) - f(n) = 10^n(10-1) + 3 \cdot 4^{n+2}(4-1)\)M1
\(= 9(10^n + 4^{n+2})\)
\(= 9\beta\)A1
Hence \(f(n+1) = 9(\beta + \alpha) \Rightarrow H_{k+1}\) is trueA1
\(H_1\) is true and \(H_k \Rightarrow H_{k+1}\), hence by PMI \(H_n\) is true for all positive integers \(n\)A1 [6] N.B. Or can show \(f(n+1) = 9(10\alpha - 2 \cdot 4^{n+2} - 5)\) for M1A1A1 (3rd, 4th & 5th marks) 
## Question 3:

| Working/Answer | Marks | Guidance |
|---|---|---|
| For $n=1$: $10 + 192 + 5 = 207 = 9 \times 23 \Rightarrow H_1$ is true | B1 | |
| Assume $H_k$ is true for some positive integer $k \Rightarrow 10^n + 3 \cdot 4^{n+2} + 5 = 9\alpha$ | B1 | |
| Hence $f(n+1) - f(n) = 10^n(10-1) + 3 \cdot 4^{n+2}(4-1)$ | M1 | |
| $= 9(10^n + 4^{n+2})$ | | |
| $= 9\beta$ | A1 | |
| Hence $f(n+1) = 9(\beta + \alpha) \Rightarrow H_{k+1}$ is true | A1 | |
| $H_1$ is true and $H_k \Rightarrow H_{k+1}$, hence by PMI $H_n$ is true for all positive integers $n$ | A1 [6] | N.B. Or can show $f(n+1) = 9(10\alpha - 2 \cdot 4^{n+2} - 5)$ for **M1A1A1** (3rd, 4th & 5th marks) |

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3 Prove by mathematical induction that, for all positive integers $n , 10 ^ { n } + 3 \times 4 ^ { n + 2 } + 5$ is divisible by 9 .

\hfill \mbox{\textit{CAIE FP1 2016 Q3 [6]}}
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