| Exam Board | Edexcel |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2012 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sequences and series, recurrence and convergence |
| Type | Sum from n+1 to 2n or similar range |
| Difficulty | Challenging +1.2 This is a structured Further Maths question with clear scaffolding through parts (a)-(c). Part (a) is routine partial fractions, part (b) is a standard method of differences proof (common FP2 technique), and part (c) requires recognizing that the sum from n+1 to 2n equals S(2n) - S(n), then algebraic manipulation. While it requires multiple steps and careful algebra, the techniques are all standard FP2 material with no novel insight needed. |
| Spec | 4.06b Method of differences: telescoping series |
\begin{enumerate}
\item (a) Express $\frac { 1 } { r ( r + 2 ) }$ in partial fractions.\\
(b) Hence prove, by the method of differences, that
\end{enumerate}
$$\sum _ { r = 1 } ^ { n } \frac { 1 } { r ( r + 2 ) } = \frac { n ( a n + b ) } { 4 ( n + 1 ) ( n + 2 ) }$$
where $a$ and $b$ are constants to be found.\\
(c) Hence show that
$$\sum _ { r = n + 1 } ^ { 2 n } \frac { 1 } { r ( r + 2 ) } = \frac { n ( 4 n + 5 ) } { 4 ( n + 1 ) ( n + 2 ) ( 2 n + 1 ) }$$
\hfill \mbox{\textit{Edexcel FP2 2012 Q6 [11]}}