| Exam Board | AQA |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2011 |
| Session | June |
| Marks | 6 |
| Paper | 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 algebraic manipulation of factorials and telescoping series. Part (a) is straightforward factorization, but part (b) requires recognizing the telescoping pattern from part (a) and carefully tracking terms—a technique beyond standard A-level. The multi-step reasoning and series manipulation place it moderately above average difficulty. |
| Spec | 4.06b Method of differences: telescoping series |
\begin{enumerate}[label=(\alph*)]
\item Show that
$$(r + 1)! - (r - 1)! = (r^2 + r - 1)(r - 1)!$$ [2 marks]
\item Hence show that
$$\sum_{r=1}^{n} (r^2 + r - 1)(r - 1)! = (n + 2)n! - 2$$ [4 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA FP2 2011 Q3 [6]}}