CAIE FP1 2014 June — Question 3 7 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2014
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicProof by induction
TypeProve divisibility
DifficultyStandard +0.3 This is a straightforward proof by induction for divisibility. The structure is standard: verify base case n=1, assume for n=k, prove for n=k+1. The algebraic manipulation required (factoring out 5^k and showing the remainder is divisible by 8) is routine for Further Maths students who have practiced induction. While it's a Further Maths topic, it's a textbook example with no novel insight required, making it slightly easier than average overall.
Spec4.01a Mathematical induction: construct proofs4.05c Partial fractions: extended to quadratic denominators

3
- 2
0 \end{array} \right) .$$ Show that \(\{ \mathbf { a } , \mathbf { b } , \mathbf { c } \}\) is a basis for \(\mathbb { R } ^ { 3 }\). Express \(\mathbf { d }\) in terms of \(\mathbf { a } , \mathbf { b }\) and \(\mathbf { c }\). 2 Show that the difference between the squares of consecutive integers is an odd integer. Find the sum to \(n\) terms of the series $$\frac { 3 } { 1 ^ { 2 } \times 2 ^ { 2 } } + \frac { 5 } { 2 ^ { 2 } \times 3 ^ { 2 } } + \frac { 7 } { 3 ^ { 2 } \times 4 ^ { 2 } } + \ldots + \frac { 2 r + 1 } { r ^ { 2 } ( r + 1 ) ^ { 2 } } + \ldots$$ and deduce the sum to infinity of the series. 3 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 3:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\phi(1)=5\times5-1=24\) which is divisible by \(8\Rightarrow H_1\) is trueB1
Assume \(P_k\) is true for some positive integer \(k\Rightarrow\phi(k)=8l\)B1
\(\phi(k+1)-\phi(k)=5^{k+1}(4k+5)-1-5^k(4k+1)+1\)M1
\(=5^k(20k+25-4k-1)\)A1
\(=5^k(16k+24)=8m\)A1
\(\therefore\phi(k+1)=8(l+m)\)A1
Hence, by PMI, true for all positive integers \(n\)A1 [7] CWO – all previous marks required
Alternative: \(\phi(k+1)=5^{k+1}(4k+5)-1=5.(4k.5^k)+25.5^k-1\)(M1A1)
\(=5(8l-5^k+1)+25.5^k-1=40l+20.5^k+4=40l+24.5^k-4.5^k+4\)(A1)
\(=40l+24.5^k-4(5^k-1)=40l+24.5^k-4(8l-4k.5^k)=8l+24.5^k+16k.5^k=8m\)(A1)
## Question 3:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\phi(1)=5\times5-1=24$ which is divisible by $8\Rightarrow H_1$ is true | B1 | |
| Assume $P_k$ is true for some positive integer $k\Rightarrow\phi(k)=8l$ | B1 | |
| $\phi(k+1)-\phi(k)=5^{k+1}(4k+5)-1-5^k(4k+1)+1$ | M1 | |
| $=5^k(20k+25-4k-1)$ | A1 | |
| $=5^k(16k+24)=8m$ | A1 | |
| $\therefore\phi(k+1)=8(l+m)$ | A1 | |
| Hence, by PMI, true for all positive integers $n$ | A1 | [7] CWO – all previous marks required |
| **Alternative:** $\phi(k+1)=5^{k+1}(4k+5)-1=5.(4k.5^k)+25.5^k-1$ | (M1A1) | |
| $=5(8l-5^k+1)+25.5^k-1=40l+20.5^k+4=40l+24.5^k-4.5^k+4$ | (A1) | |
| $=40l+24.5^k-4(5^k-1)=40l+24.5^k-4(8l-4k.5^k)=8l+24.5^k+16k.5^k=8m$ | (A1) | |

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3 \\
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0
\end{array} \right) .$$

Show that $\{ \mathbf { a } , \mathbf { b } , \mathbf { c } \}$ is a basis for $\mathbb { R } ^ { 3 }$.

Express $\mathbf { d }$ in terms of $\mathbf { a } , \mathbf { b }$ and $\mathbf { c }$.

2 Show that the difference between the squares of consecutive integers is an odd integer.

Find the sum to $n$ terms of the series

$$\frac { 3 } { 1 ^ { 2 } \times 2 ^ { 2 } } + \frac { 5 } { 2 ^ { 2 } \times 3 ^ { 2 } } + \frac { 7 } { 3 ^ { 2 } \times 4 ^ { 2 } } + \ldots + \frac { 2 r + 1 } { r ^ { 2 } ( r + 1 ) ^ { 2 } } + \ldots$$

and deduce the sum to infinity of the series.

3 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 FP1 2014 Q3 [7]}}