| Exam Board | Edexcel |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2013 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Partial Fractions |
| Type | Two linear factors in denominator |
| Difficulty | Moderate -0.3 This is a straightforward partial fractions question with simple linear factors and a standard telescoping series. Part (a) is routine algebraic manipulation, and part (b) requires recognizing the telescoping pattern—a common technique at this level. While it's from FP2, the question is more mechanical than conceptually challenging, making it slightly easier than average. |
| Spec | 4.06b Method of differences: telescoping series |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{2}{(2r+1)(2r+3)} = \frac{A}{2r+1} + \frac{B}{2r+3}\) | M1 | For any valid attempt to obtain the partial fractions |
| \(= \frac{1}{2r+1} - \frac{1}{2r+3}\) | A1 | For \(\frac{1}{2r+1} - \frac{1}{2r+3}\) specifically |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{1}{3} - \frac{1}{5} + \frac{1}{5} - \frac{1}{7} + \cdots + \frac{1}{2n+1} - \frac{1}{2n+3}\) | M1 | For using their PFs to split each term; at least 2 terms at start and 1 at end needed to show diagonal cancellation resulting in two remaining terms |
| \(= \frac{1}{3} - \frac{1}{2n+3} = \frac{2n+3-3}{3(2n+3)}\) | M1dep | For simplifying to a single fraction and multiplying by the appropriate constant |
| \(\displaystyle\sum_{1}^{n} \frac{3}{(2r+1)(2r+3)} = \frac{3}{2} \times \frac{2n}{3(2n+3)} = \frac{n}{2n+3}\) | A1cao | For \(\displaystyle\sum = \frac{n}{2n+3}\) |
## Question 1:
### Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{2}{(2r+1)(2r+3)} = \frac{A}{2r+1} + \frac{B}{2r+3}$ | M1 | For any valid attempt to obtain the partial fractions |
| $= \frac{1}{2r+1} - \frac{1}{2r+3}$ | A1 | For $\frac{1}{2r+1} - \frac{1}{2r+3}$ specifically |
**NB:** With no working shown, award M1A1 if correct PFs written down, but M0A0 if either one is incorrect.
**(2 marks)**
---
### Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{1}{3} - \frac{1}{5} + \frac{1}{5} - \frac{1}{7} + \cdots + \frac{1}{2n+1} - \frac{1}{2n+3}$ | M1 | For using **their** PFs to split each term; at least 2 terms at start and 1 at end needed to show diagonal cancellation resulting in two remaining terms |
| $= \frac{1}{3} - \frac{1}{2n+3} = \frac{2n+3-3}{3(2n+3)}$ | M1dep | For simplifying to a single fraction and multiplying by the appropriate constant |
| $\displaystyle\sum_{1}^{n} \frac{3}{(2r+1)(2r+3)} = \frac{3}{2} \times \frac{2n}{3(2n+3)} = \frac{n}{2n+3}$ | A1cao | For $\displaystyle\sum = \frac{n}{2n+3}$ |
**NB:** If $r$ is used instead of $n$ (including for the answer), only M marks are available.
**(3 marks)**
**[Total: 5 marks]**\begin{enumerate}
\item (a) Express $\frac { 2 } { ( 2 r + 1 ) ( 2 r + 3 ) }$ in partial fractions.\\
(b) Using your answer to (a), find, in terms of $n$,
\end{enumerate}
$$\sum _ { r = 1 } ^ { n } \frac { 3 } { ( 2 r + 1 ) ( 2 r + 3 ) }$$
Give your answer as a single fraction in its simplest form.\\
\hfill \mbox{\textit{Edexcel FP2 2013 Q1 [5]}}