| Exam Board | OCR MEI |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2008 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sequences and series, recurrence and convergence |
| Type | Method of differences with given identity |
| Difficulty | Standard +0.3 This is a straightforward application of the method of differences with the identity provided. Part (i) is routine algebraic verification requiring only finding a common denominator. Part (ii) is a standard textbook exercise where the telescoping pattern is clear once the identity is applied. While it's a Further Maths topic, the execution requires no novel insight and follows a well-practiced technique, making it slightly easier than average overall. |
| Spec | 4.06b Method of differences: telescoping series |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{1}{3r-1} - \frac{1}{3r+2} \equiv \frac{3r+2-(3r-1)}{(3r-1)(3r+2)} \equiv \frac{3}{(3r-1)(3r+2)}\) | M1, A1 [2] | Attempt at correct method; Correct, without fudging |
| Answer | Marks | Guidance |
|---|---|---|
| \(\sum_{r=1}^{n} \frac{1}{(3r-1)(3r+2)} = \frac{1}{3}\sum_{r=1}^{n}\left[\frac{1}{3r-1} - \frac{1}{3r+2}\right]\) | M1 | Attempt to use identity |
| \(= \frac{1}{3}\left[\left(\frac{1}{2}-\frac{1}{5}\right)+\left(\frac{1}{5}-\frac{1}{8}\right)+\cdots+\left(\frac{1}{3n-1}-\frac{1}{3n+2}\right)\right]\) | A1, M1 | Terms in full (at least two); Attempt at cancelling |
| \(= \frac{1}{3}\left[\frac{1}{2} - \frac{1}{3n+2}\right]\) | A2 [5] | A1 if factor of \(\frac{1}{3}\) missing; A1 max if answer not in terms of \(n\) |
# Question 7:
## Question 7(i):
| $\frac{1}{3r-1} - \frac{1}{3r+2} \equiv \frac{3r+2-(3r-1)}{(3r-1)(3r+2)} \equiv \frac{3}{(3r-1)(3r+2)}$ | M1, A1 [2] | Attempt at correct method; Correct, without fudging |
## Question 7(ii):
| $\sum_{r=1}^{n} \frac{1}{(3r-1)(3r+2)} = \frac{1}{3}\sum_{r=1}^{n}\left[\frac{1}{3r-1} - \frac{1}{3r+2}\right]$ | M1 | Attempt to use identity |
| $= \frac{1}{3}\left[\left(\frac{1}{2}-\frac{1}{5}\right)+\left(\frac{1}{5}-\frac{1}{8}\right)+\cdots+\left(\frac{1}{3n-1}-\frac{1}{3n+2}\right)\right]$ | A1, M1 | Terms in full (at least two); Attempt at cancelling |
| $= \frac{1}{3}\left[\frac{1}{2} - \frac{1}{3n+2}\right]$ | A2 [5] | A1 if factor of $\frac{1}{3}$ missing; A1 max if answer not in terms of $n$ |
---7 (i) Show that $\frac { 1 } { 3 r - 1 } - \frac { 1 } { 3 r + 2 } \equiv \frac { 3 } { ( 3 r - 1 ) ( 3 r + 2 ) }$ for all integers $r$.\\
(ii) Hence use the method of differences to find $\sum _ { r = 1 } ^ { n } \frac { 1 } { ( 3 r - 1 ) ( 3 r + 2 ) }$.
Section B (36 marks)\\
\hfill \mbox{\textit{OCR MEI FP1 2008 Q7 [7]}}