| Exam Board | AQA |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2011 |
| Session | January |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Topic | Sequences and series, recurrence and convergence |
| Type | Method of differences with given identity |
| Difficulty | Standard +0.3 This is a straightforward method of differences question with the identity provided. Part (a) requires algebraic expansion and simplification to verify the given result. Part (b) is a direct application of telescoping series, requiring students to recognize that the sum collapses to u_100 - u_0. While it involves Further Maths content, the technique is mechanical once the identity is given, making it slightly easier than average. |
| Spec | 4.06b Method of differences: telescoping series |
2
\begin{enumerate}[label=(\alph*)]
\item Given that
$$u _ { r } = \frac { 1 } { 6 } r ( r + 1 ) ( 4 r + 11 )$$
show that
$$u _ { r } - u _ { r - 1 } = r ( 2 r + 3 )$$
\item Hence find the sum of the first hundred terms of the series
$$1 \times 5 + 2 \times 7 + 3 \times 9 + \ldots + r ( 2 r + 3 ) + \ldots$$
\end{enumerate}
\hfill \mbox{\textit{AQA FP2 2011 Q2 [6]}}