| Exam Board | AQA |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2015 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sequences and series, recurrence and convergence |
| Type | Factorial or product method of differences |
| Difficulty | Standard +0.8 This is a Further Maths FP2 question requiring partial fraction decomposition with factorials (non-standard form) followed by telescoping series summation. Part (a) requires algebraic manipulation with factorials to find constants A and B, while part (b) applies the method of differences. The factorial context makes it less routine than standard partial fractions, but it's a recognizable FP2 technique with clear structure and only 5 marks total. |
| Spec | 4.06b Method of differences: telescoping series |
**(a) [3 marks]**
Express in the form $\frac{A}{B}$, where $A$ and $B$ are integers.
$$\frac{1}{(r+2)r!} - \frac{1}{(r+1)!} - \frac{1}{(r+2)!}$$
**(b) [2 marks]**
Hence find $\sum_{r=1}^{n} \frac{1}{(r+2)r!}$1
\begin{enumerate}[label=(\alph*)]
\item Express $\frac { 1 } { ( r + 2 ) r ! }$ in the form $\frac { A } { ( r + 1 ) ! } + \frac { B } { ( r + 2 ) ! }$, where $A$ and $B$ are integers.\\[0pt]
[3 marks]
\item Hence find $\sum _ { r = 1 } ^ { n } \frac { 1 } { ( r + 2 ) r ! }$.\\[0pt]
[2 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA FP2 2015 Q1 [5]}}