| Exam Board | Edexcel |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2010 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Partial Fractions |
| Type | Two linear factors in denominator |
| Difficulty | Standard +0.3 This is a standard Further Pure 2 partial fractions question with method of differences. Part (a) is routine decomposition with two linear factors, part (b) follows a well-practiced telescoping series technique, and part (c) is straightforward application. While it's Further Maths content, it requires no novel insight—just executing familiar procedures correctly. |
| Spec | 1.02y Partial fractions: decompose rational functions4.06b Method of differences: telescoping series |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{1}{3r-1} - \frac{1}{3r+2}\) | M1 A1 | (2 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\sum_{r=1}^{n} \frac{3}{(3r-1)(3r+2)} = \frac{1}{2} - \frac{1}{5} + \frac{1}{5} - \frac{1}{8} + \frac{1}{8} - \frac{1}{11} + \ldots + \frac{1}{3n-1} - \frac{1}{3n+2}\) | M1 A1ft | Telescoping series shown |
| \(= \frac{1}{2} - \frac{1}{3n+2} = \frac{3n}{2(3n+2)}\) * | A1 | (3 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Sum \(= f(1000) - f(99)\) | M1 | |
| \(\frac{3000}{6004} - \frac{297}{598} = 0.00301\) or \(3.01 \times 10^{-3}\) | A1 | (2 marks) |
## Question 1:
**Part (a)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{1}{3r-1} - \frac{1}{3r+2}$ | M1 A1 | (2 marks) |
**Part (b)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\sum_{r=1}^{n} \frac{3}{(3r-1)(3r+2)} = \frac{1}{2} - \frac{1}{5} + \frac{1}{5} - \frac{1}{8} + \frac{1}{8} - \frac{1}{11} + \ldots + \frac{1}{3n-1} - \frac{1}{3n+2}$ | M1 A1ft | Telescoping series shown |
| $= \frac{1}{2} - \frac{1}{3n+2} = \frac{3n}{2(3n+2)}$ * | A1 | (3 marks) |
**Part (c)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Sum $= f(1000) - f(99)$ | M1 | |
| $\frac{3000}{6004} - \frac{297}{598} = 0.00301$ or $3.01 \times 10^{-3}$ | A1 | (2 marks) |
---\begin{enumerate}
\item (a) Express $\frac { 3 } { ( 3 r - 1 ) ( 3 r + 2 ) }$ in partial fractions.\\
(b) Using your answer to part (a) and the method of differences, show that
\end{enumerate}
$$\sum _ { r = 1 } ^ { n } \frac { 3 } { ( 3 r - 1 ) ( 3 r + 2 ) } = \frac { 3 n } { 2 ( 3 n + 2 ) }$$
(c) Evaluate $\sum _ { r = 100 } ^ { 1000 } \frac { 3 } { ( 3 r - 1 ) ( 3 r + 2 ) }$, giving your answer to 3 significant figures.\\
\hfill \mbox{\textit{Edexcel FP2 2010 Q1 [7]}}