| Exam Board | CAIE |
|---|---|
| Module | Further Paper 1 (Further Paper 1) |
| Year | 2021 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sequences and series, recurrence and convergence |
| Type | Method of differences with logarithmic terms |
| Difficulty | Challenging +1.2 This is a Further Maths question requiring method of differences with logarithms, which is more sophisticated than standard A-level. However, the telescoping pattern is relatively straightforward once the logarithm is split using log laws, and part (b) is a routine application of the result from part (a). The question requires solid technique but no particularly novel insight. |
| Spec | 4.06b Method of differences: telescoping series |
| Answer | Marks | Guidance |
|---|---|---|
| \(\sum_{r=1}^{n}(\ln r - 2\ln(r+1) + \ln(r+2))\) | B1 | Separates logarithms into correct form using a difference, or as logarithm of product |
| \(\ln 1 - 2\ln 2 + \ln 3\)<br>\(\ln 2 - 2\ln 3 + \ln 4\)<br>\(\ln 3 - 2\ln 4 + \ln 5\)<br>\(\vdots\)<br>\(\ln(n-1) - 2\ln n + \ln(n+1)\)<br>\(\ln n - 2\ln(n+1) + \ln(n+2)\) | M1 A1 | Shows enough terms to make cancellation clear |
| \([\ln 1] - \ln 2 - \ln(n+1) + \ln(n+2) = \ln\dfrac{n+2}{2(n+1)}\) | A1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| \(S = -\ln 2\) | B1 | States sum to infinity. AEF |
| \(S_n - S = \ln\left(\dfrac{n+2}{n+1}\right) < 0.01\) leading to \(n+2 < e^{0.01}(n+1)\) | M1 | Forms inequality |
| Least value of \(n\) is 99 | A1 | CAO |
## Question 3(a):
| $\sum_{r=1}^{n}(\ln r - 2\ln(r+1) + \ln(r+2))$ | B1 | Separates logarithms into correct form using a difference, or as logarithm of product |
|---|---|---|
| $\ln 1 - 2\ln 2 + \ln 3$<br>$\ln 2 - 2\ln 3 + \ln 4$<br>$\ln 3 - 2\ln 4 + \ln 5$<br>$\vdots$<br>$\ln(n-1) - 2\ln n + \ln(n+1)$<br>$\ln n - 2\ln(n+1) + \ln(n+2)$ | M1 A1 | Shows enough terms to make cancellation clear |
| $[\ln 1] - \ln 2 - \ln(n+1) + \ln(n+2) = \ln\dfrac{n+2}{2(n+1)}$ | A1 | AG |
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## Question 3(b):
| $S = -\ln 2$ | B1 | States sum to infinity. AEF |
|---|---|---|
| $S_n - S = \ln\left(\dfrac{n+2}{n+1}\right) < 0.01$ leading to $n+2 < e^{0.01}(n+1)$ | M1 | Forms inequality |
| Least value of $n$ is 99 | A1 | CAO |
---3 Let $S _ { n } = \sum _ { r = 1 } ^ { n } \ln \frac { r ( r + 2 ) } { ( r + 1 ) ^ { 2 } }$.
\begin{enumerate}[label=(\alph*)]
\item Using the method of differences, or otherwise, show that $S _ { n } = \ln \frac { n + 2 } { 2 ( n + 1 ) }$.\\
Let $S = \sum _ { r = 1 } ^ { \infty } \ln \frac { r ( r + 2 ) } { ( r + 1 ) ^ { 2 } }$.
\item Find the least value of $n$ such that $\mathrm { S } _ { \mathrm { n } } - \mathrm { S } < 0.01$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 1 2021 Q3 [7]}}