| Exam Board | Edexcel |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2009 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Partial Fractions |
| Type | Two linear factors in denominator |
| Difficulty | Standard +0.3 This is a straightforward Further Maths question with two standard parts: (a) routine partial fractions decomposition with simple linear factors, and (b) telescoping series summation requiring algebraic manipulation but following a well-established method. While it's FP2, both techniques are mechanical and commonly practiced, making it slightly easier than average overall. |
| Spec | 1.02y Partial fractions: decompose rational functions4.06b Method of differences: telescoping series |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{1}{r(r+2)} = \frac{1}{2r} - \frac{1}{2(r+2)}\) | B1 aef | (1 mark) |
| Answer | Marks | Guidance |
|---|---|---|
| List first two terms and last two terms: \(\left(\frac{2}{1}-\frac{2}{3}\right)+\left(\frac{2}{2}-\frac{2}{4}\right)+\ldots+\left(\frac{2}{n-1}-\frac{2}{n+1}\right)+\left(\frac{2}{n}-\frac{2}{n+2}\right)\) | M1 | List first two terms and last two terms |
| \(= \frac{2}{1}+\frac{2}{2}-\frac{2}{n+1}-\frac{2}{n+2}\) | M1, A1 | Includes first two and final two underlined terms; \(\frac{2}{1}+\frac{2}{2}-\frac{2}{n+1}-\frac{2}{n+2}\) |
| \(= 3 - \frac{2}{n+1} - \frac{2}{n+2}\) | ||
| \(= \frac{3(n+1)(n+2)-2(n+2)-2(n+1)}{(n+1)(n+2)}\) | M1 | Attempt to combine at least 3-term fraction to single fraction and attempt to expand brackets from numerator |
| \(= \frac{3n^2+9n+6-2n-4-2n-2}{(n+1)(n+2)}\) | ||
| \(= \frac{3n^2+5n}{(n+1)(n+2)}\) | ||
| \(= \frac{n(3n+5)}{(n+1)(n+2)}\) | A1 cso AG | Correct result (5 marks) |
# Question 1:
## Part (a)
| $\frac{1}{r(r+2)} = \frac{1}{2r} - \frac{1}{2(r+2)}$ | B1 **aef** | (1 mark) |
## Part (b)
| List first two terms and last two terms: $\left(\frac{2}{1}-\frac{2}{3}\right)+\left(\frac{2}{2}-\frac{2}{4}\right)+\ldots+\left(\frac{2}{n-1}-\frac{2}{n+1}\right)+\left(\frac{2}{n}-\frac{2}{n+2}\right)$ | M1 | List first two terms and last two terms |
|---|---|---|
| $= \frac{2}{1}+\frac{2}{2}-\frac{2}{n+1}-\frac{2}{n+2}$ | M1, A1 | Includes first two and final two underlined terms; $\frac{2}{1}+\frac{2}{2}-\frac{2}{n+1}-\frac{2}{n+2}$ |
| $= 3 - \frac{2}{n+1} - \frac{2}{n+2}$ | | |
| $= \frac{3(n+1)(n+2)-2(n+2)-2(n+1)}{(n+1)(n+2)}$ | M1 | Attempt to combine at least 3-term fraction to single fraction and attempt to expand brackets from numerator |
| $= \frac{3n^2+9n+6-2n-4-2n-2}{(n+1)(n+2)}$ | | |
| $= \frac{3n^2+5n}{(n+1)(n+2)}$ | | |
| $= \frac{n(3n+5)}{(n+1)(n+2)}$ | A1 **cso AG** | Correct result (5 marks) |
---\begin{enumerate}[label=(\alph*)]
\item Express $\frac { 1 } { r ( r + 2 ) }$ in partial fractions.
\item Hence show that $\sum _ { r = 1 } ^ { n } \frac { 4 } { r ( r + 2 ) } = \frac { n ( 3 n + 5 ) } { ( n + 1 ) ( n + 2 ) }$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP2 2009 Q1 [6]}}