| Exam Board | OCR MEI |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2009 |
| Session | June |
| Marks | 6 |
| 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 standard method of differences question with the partial fraction decomposition already provided. Part (i) requires only algebraic verification (combining fractions), and part (ii) is a routine telescoping series application. While it's a Further Maths topic, the execution is mechanical with no novel insight required, making it slightly easier than an average A-level question overall. |
| Spec | 1.02y Partial fractions: decompose rational functions4.06b Method of differences: telescoping series |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{1}{5r-2} - \frac{1}{5r+3} \equiv \frac{5r+3-5r+2}{(5r+3)(5r-2)}\) | M1 | Attempt to form common denominator |
| \(\equiv \frac{5}{(5r+3)(5r-2)}\) | A1 [2] | Correct cancelling |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\sum_{r=1}^{n} \frac{1}{(5r-2)(5r+3)} = \frac{1}{5}\sum_{r=1}^{n}\left[\frac{1}{(5r-2)} - \frac{1}{(5r+3)}\right]\) | ||
| \(= \frac{1}{5}\left[\left(\frac{1}{3}-\frac{1}{8}\right)+\left(\frac{1}{8}-\frac{1}{13}\right)+\left(\frac{1}{13}-\frac{1}{18}\right)+\cdots\right]\) | B1 | First two terms in full |
| \(+\left(\frac{1}{5n-7}-\frac{1}{5n-2}\right)+\left(\frac{1}{5n-2}-\frac{1}{5n+3}\right)\) | B1 | Last term in full |
| M1 | Attempt to cancel terms | |
| \(= \frac{1}{5}\left[\frac{1}{3} - \frac{1}{5n+3}\right] = \frac{n}{3(5n+3)}\) | A1 [4] |
# Question 5(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{1}{5r-2} - \frac{1}{5r+3} \equiv \frac{5r+3-5r+2}{(5r+3)(5r-2)}$ | M1 | Attempt to form common denominator |
| $\equiv \frac{5}{(5r+3)(5r-2)}$ | A1 **[2]** | Correct cancelling |
---
# Question 5(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\sum_{r=1}^{n} \frac{1}{(5r-2)(5r+3)} = \frac{1}{5}\sum_{r=1}^{n}\left[\frac{1}{(5r-2)} - \frac{1}{(5r+3)}\right]$ | | |
| $= \frac{1}{5}\left[\left(\frac{1}{3}-\frac{1}{8}\right)+\left(\frac{1}{8}-\frac{1}{13}\right)+\left(\frac{1}{13}-\frac{1}{18}\right)+\cdots\right]$ | B1 | First two terms in full |
| $+\left(\frac{1}{5n-7}-\frac{1}{5n-2}\right)+\left(\frac{1}{5n-2}-\frac{1}{5n+3}\right)$ | B1 | Last term in full |
| | M1 | Attempt to cancel terms |
| $= \frac{1}{5}\left[\frac{1}{3} - \frac{1}{5n+3}\right] = \frac{n}{3(5n+3)}$ | A1 **[4]** | |
---5 (i) Show that $\frac { 1 } { 5 r - 2 } - \frac { 1 } { 5 r + 3 } \equiv \frac { 5 } { ( 5 r - 2 ) ( 5 r + 3 ) }$ for all integers $r$.\\
(ii) Hence use the method of differences to show that $\sum _ { r = 1 } ^ { n } \frac { 1 } { ( 5 r - 2 ) ( 5 r + 3 ) } = \frac { n } { 3 ( 5 n + 3 ) }$.
\hfill \mbox{\textit{OCR MEI FP1 2009 Q5 [6]}}