CAIE FP1 2017 June — Question 3 6 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2017
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicProof by induction
TypeProve summation with logarithms
DifficultyStandard +0.8 This is a Further Maths induction proof combining logarithms with factorials. While the inductive step requires careful algebraic manipulation of log properties and recognizing how to simplify (n+1)ln((n+2)/(n+1)) + ln((n+1)^n/n!) into the required form, it follows a standard induction template without requiring deep insight. The 6-mark allocation and FP1 level suggest moderate difficulty above typical A-level but not exceptionally challenging.
Spec4.01a Mathematical induction: construct proofs
3 Prove, by mathematical induction, that \(\sum _ { r = 1 } ^ { n } r \ln \left( \frac { r + 1 } { r } \right) = \ln \left( \frac { ( n + 1 ) ^ { n } } { n ! } \right)\) for all positive integers \(n\). [6]
Question 3:
AnswerMarks Guidance
AnswerMarks Guidance
When \(n=1\): \(1 \times \ln 2 = \ln 2 = \ln\!\left(\frac{2^1}{1!}\right) \Rightarrow (H_1 \text{ is true})\)B1
Assume for some positive integer \(k\): \(\sum_{r=1}^{k} r\ln\!\left(\frac{r+1}{r}\right) = \ln\!\left(\frac{[k+1]^k}{k!}\right)\)B1
Hence \(\sum_{r=1}^{k+1} r\ln\!\left(\frac{r+1}{r}\right) = \ln\!\left(\frac{(k+1)^k}{k!}\right) + (k+1)\ln\!\left(\frac{k+2}{k+1}\right)\)B1
\(= \ln\frac{(k+1)^k(k+2)^{k+1}}{k!(k+1)^{k+1}}\)M1
\(= \ln\frac{(k+1)^k(k+2)^{k+1}}{(k+1)!(k+1)^k} = \ln\frac{(k+2)^{k+1}}{(k+1)!}\)A1
Thus \(H_k \Rightarrow H_{k+1}\) and hence by PMI \(H_n\) is true for all positive integersA1 
## Question 3:

| Answer | Marks | Guidance |
|--------|-------|----------|
| When $n=1$: $1 \times \ln 2 = \ln 2 = \ln\!\left(\frac{2^1}{1!}\right) \Rightarrow (H_1 \text{ is true})$ | B1 | |
| Assume for some positive integer $k$: $\sum_{r=1}^{k} r\ln\!\left(\frac{r+1}{r}\right) = \ln\!\left(\frac{[k+1]^k}{k!}\right)$ | B1 | |
| Hence $\sum_{r=1}^{k+1} r\ln\!\left(\frac{r+1}{r}\right) = \ln\!\left(\frac{(k+1)^k}{k!}\right) + (k+1)\ln\!\left(\frac{k+2}{k+1}\right)$ | B1 | |
| $= \ln\frac{(k+1)^k(k+2)^{k+1}}{k!(k+1)^{k+1}}$ | M1 | |
| $= \ln\frac{(k+1)^k(k+2)^{k+1}}{(k+1)!(k+1)^k} = \ln\frac{(k+2)^{k+1}}{(k+1)!}$ | A1 | |
| Thus $H_k \Rightarrow H_{k+1}$ and hence by PMI $H_n$ is true for all positive integers | A1 | |
3 Prove, by mathematical induction, that $\sum _ { r = 1 } ^ { n } r \ln \left( \frac { r + 1 } { r } \right) = \ln \left( \frac { ( n + 1 ) ^ { n } } { n ! } \right)$ for all positive integers $n$. [6]\\

\hfill \mbox{\textit{CAIE FP1 2017 Q3 [6]}}
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