| Exam Board | OCR MEI |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2005 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sequences and series, recurrence and convergence |
| Type | Infinite series convergence and sum |
| Difficulty | Standard +0.3 This is a guided method of differences question where the partial fraction decomposition is given. Part (i) requires writing out terms and identifying telescoping cancellations (standard FP1 technique), and part (ii) is a direct application taking the limit as nāā. The question is slightly easier than average because the decomposition is provided and the method is explicitly named, requiring only careful execution rather than insight. |
| Spec | 4.06b Method of differences: telescoping series |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\sum_{r=1}^{n} \frac{2}{r(r+1)(r+2)} = \sum_{r=1}^{n} \left[\frac{1}{r} - \frac{2}{(r+1)} + \frac{1}{(r+2)}\right]\) | M1 | Give if implied by later working |
| Writing out at least three terms in full: \(\left(\frac{1}{1}-\frac{2}{2}+\frac{1}{3}\right)+\left(\frac{1}{2}-\frac{2}{3}+\frac{1}{4}\right)+\left(\frac{1}{3}-\frac{2}{4}+\frac{1}{5}\right)+\cdots+\left(\frac{1}{n-1}-\frac{2}{n}+\frac{1}{n+1}\right)+\left(\frac{1}{n}-\frac{2}{n+1}+\frac{1}{n+2}\right)\) | M1 A2 | Writing out terms in full, at least three terms. All terms correct. A1 for at least two correct |
| \(= \frac{1}{1}-\frac{2}{2}+\frac{1}{2}+\frac{1}{n+1}-\frac{2}{n+1}+\frac{1}{n+2}\) | M1 A3 | Attempt at cancelling terms. Correct terms retained (minus 1 each error) |
| \(= \frac{1}{2} - \frac{1}{n+1} + \frac{1}{n+2}\) | ||
| \(= \frac{1}{2} - \frac{1}{(n+1)(n+2)}\) | M1 | Attempt at single fraction leading to given answer |
| [9 marks] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{1}{1\times2\times3}+\frac{1}{2\times3\times4}+\frac{1}{3\times4\times5}+\cdots\) | ||
| \(= \frac{1}{2}\sum_{r=1}^{n}\frac{2}{r(r+1)(r+2)} = \frac{1}{2}\left(\frac{1}{2}-\frac{1}{(n+1)(n+2)}\right)\) | M1 | Relating to previous sum |
| As \(n\to\infty\), \(\frac{1}{(n+1)(n+2)}\to 0\) | M1 | Recognising that \(\frac{1}{(n+1)(n+2)}\to 0\) as \(n\to\infty\) (could be implied) |
| \(\Rightarrow \frac{1}{1\times2\times3}+\frac{1}{2\times3\times4}+\frac{1}{3\times4\times5}+\cdots = \frac{1}{4}\) | A1 | |
| [3 marks] |
## Question 10:
### Part 10(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\sum_{r=1}^{n} \frac{2}{r(r+1)(r+2)} = \sum_{r=1}^{n} \left[\frac{1}{r} - \frac{2}{(r+1)} + \frac{1}{(r+2)}\right]$ | M1 | Give if implied by later working |
| Writing out at least three terms in full: $\left(\frac{1}{1}-\frac{2}{2}+\frac{1}{3}\right)+\left(\frac{1}{2}-\frac{2}{3}+\frac{1}{4}\right)+\left(\frac{1}{3}-\frac{2}{4}+\frac{1}{5}\right)+\cdots+\left(\frac{1}{n-1}-\frac{2}{n}+\frac{1}{n+1}\right)+\left(\frac{1}{n}-\frac{2}{n+1}+\frac{1}{n+2}\right)$ | M1 A2 | Writing out terms in full, at least three terms. All terms correct. A1 for at least two correct |
| $= \frac{1}{1}-\frac{2}{2}+\frac{1}{2}+\frac{1}{n+1}-\frac{2}{n+1}+\frac{1}{n+2}$ | M1 A3 | Attempt at cancelling terms. Correct terms retained (minus 1 each error) |
| $= \frac{1}{2} - \frac{1}{n+1} + \frac{1}{n+2}$ | | |
| $= \frac{1}{2} - \frac{1}{(n+1)(n+2)}$ | M1 | Attempt at single fraction leading to given answer |
| **[9 marks]** | | |
---
### Part 10(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{1}{1\times2\times3}+\frac{1}{2\times3\times4}+\frac{1}{3\times4\times5}+\cdots$ | | |
| $= \frac{1}{2}\sum_{r=1}^{n}\frac{2}{r(r+1)(r+2)} = \frac{1}{2}\left(\frac{1}{2}-\frac{1}{(n+1)(n+2)}\right)$ | M1 | Relating to previous sum |
| As $n\to\infty$, $\frac{1}{(n+1)(n+2)}\to 0$ | M1 | Recognising that $\frac{1}{(n+1)(n+2)}\to 0$ as $n\to\infty$ (could be implied) |
| $\Rightarrow \frac{1}{1\times2\times3}+\frac{1}{2\times3\times4}+\frac{1}{3\times4\times5}+\cdots = \frac{1}{4}$ | A1 | |
| **[3 marks]** | | |10 (i) You are given that
$$\frac { 2 } { r ( r + 1 ) ( r + 2 ) } = \frac { 1 } { r } - \frac { 2 } { r + 1 } + \frac { 1 } { r + 2 }$$
Use the method of differences to show that
$$\sum _ { r = 1 } ^ { n } \frac { 2 } { r ( r + 1 ) ( r + 2 ) } = \frac { 1 } { 2 } - \frac { 1 } { ( n + 1 ) ( n + 2 ) }$$
(ii) Hence find the sum of the infinite series
$$\frac { 1 } { 1 \times 2 \times 3 } + \frac { 1 } { 2 \times 3 \times 4 } + \frac { 1 } { 3 \times 4 \times 5 } + \ldots$$
\hfill \mbox{\textit{OCR MEI FP1 2005 Q10 [12]}}