| Exam Board | Edexcel |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2013 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Partial Fractions |
| Type | Three linear factors in denominator |
| Difficulty | Standard +0.8 This is a standard Further Maths partial fractions question with three linear factors, followed by a telescoping series proof. Part (a) is routine FP2 content, but part (b) requires careful bookkeeping with the method of differences and algebraic manipulation to reach the given form. The multi-step nature and need to recognize the telescoping pattern elevates this above average difficulty. |
| Spec | 4.06b Method of differences: telescoping series |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{1}{r} - \frac{2}{(r+1)} + \frac{1}{(r+2)}\) | M1 A1 A1 | (3 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Write out LHS as telescoping sum beginning \(\frac{1}{1} - \frac{2}{2} + \frac{1}{3} + \frac{1}{2} - \frac{2}{3} + \frac{1}{4} + \ldots\) | M1 | |
| Complete final three rows shown: \(+\frac{1}{(n-2)} - \frac{2}{(n-1)} + \frac{1}{n}\), \(+\frac{1}{(n-1)} - \frac{2}{n} + \frac{1}{(n+1)}\), \(+\frac{1}{n} - \frac{2}{(n+1)} + \frac{1}{(n+2)}\) | A1 | |
| \(= 1 - \frac{1}{2} - \frac{1}{n+1} + \frac{1}{n+2} = \frac{n^2+3n+2-2n-4+2n+2}{2(n+1)(n+2)}\) | M1 dep | |
| \(= \frac{n^2+3n}{2(n+1)(n+2)} = \frac{n(n+3)}{2(n+1)(n+2)}\) * | A1 | (4 marks) |
## Question 5:
### Part (a):
| $\frac{1}{r} - \frac{2}{(r+1)} + \frac{1}{(r+2)}$ | M1 A1 A1 | (3 marks) |
### Part (b):
| Write out LHS as telescoping sum beginning $\frac{1}{1} - \frac{2}{2} + \frac{1}{3} + \frac{1}{2} - \frac{2}{3} + \frac{1}{4} + \ldots$ | M1 | |
| Complete final three rows shown: $+\frac{1}{(n-2)} - \frac{2}{(n-1)} + \frac{1}{n}$, $+\frac{1}{(n-1)} - \frac{2}{n} + \frac{1}{(n+1)}$, $+\frac{1}{n} - \frac{2}{(n+1)} + \frac{1}{(n+2)}$ | A1 | |
| $= 1 - \frac{1}{2} - \frac{1}{n+1} + \frac{1}{n+2} = \frac{n^2+3n+2-2n-4+2n+2}{2(n+1)(n+2)}$ | M1 dep | |
| $= \frac{n^2+3n}{2(n+1)(n+2)} = \frac{n(n+3)}{2(n+1)(n+2)}$ * | A1 | (4 marks) |
---5.\\
\begin{enumerate}[label=(\alph*)]
\item Express $\frac { 2 } { r ( r + 1 ) ( r + 2 ) }$ in partial fractions.
\item Using your answer to part (a) and the method of differences, show that
$$\sum _ { r = 1 } ^ { n } \frac { 2 } { r ( r + 1 ) ( r + 2 ) } = \frac { n ( n + 3 ) } { 2 ( n + 1 ) ( n + 2 ) }$$
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP2 2013 Q5 [7]}}