| Exam Board | Edexcel |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2014 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sequences and series, recurrence and convergence |
| Type | Method of differences with given identity |
| Difficulty | Standard +0.8 This is a standard Further Maths method of differences question requiring partial fractions decomposition followed by telescoping series summation. While it involves multiple steps and FP2 content, it follows a well-established template with no novel insight required—students who have practiced this technique will find it routine. |
| Spec | 4.06b Method of differences: telescoping series |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{2}{4r^2-1} = \frac{A}{2r+1} + \frac{B}{2r-1}\) | ||
| \(2 = A(2r-1) + B(2r+1) \Rightarrow A = -1, B = 1\) | M1 | Complete method for finding PFs |
| \(\frac{2}{4r^2-1} = \frac{1}{2r-1} - \frac{1}{2r+1}\) | A1 | Both PFs correct; award M1A1 for both PFs seen correct without working, M0A0 otherwise |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\sum_{r=1}^{n} \frac{1}{4r^2-1} = \sum_{r=1}^{n}\left(\frac{1}{2r-1} - \frac{1}{2r+1}\right)\) | ||
| \(= 1 - \frac{1}{3} + \frac{1}{3} - \frac{1}{5} + \frac{1}{5} - \frac{1}{7} + \ldots + \frac{1}{2n-1} - \frac{1}{2n+1} = 1 - \frac{1}{2n+1}\) | M1A1ft | M1: showing fractions with PFs, min 2 at start and 1 at end, must start at 1 and end at \(n\); A1ft: identify 2 non-cancelling fractions |
| \(= \frac{2n+1-1}{2n+1}\) | ||
| \(\sum_{r=1}^{n} \frac{1}{4r^2-1} = \frac{n}{2n+1}\) | A1 | A1cso: correct final answer |
# Question 1:
## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{2}{4r^2-1} = \frac{A}{2r+1} + \frac{B}{2r-1}$ | | |
| $2 = A(2r-1) + B(2r+1) \Rightarrow A = -1, B = 1$ | M1 | Complete method for finding PFs |
| $\frac{2}{4r^2-1} = \frac{1}{2r-1} - \frac{1}{2r+1}$ | A1 | Both PFs correct; award M1A1 for both PFs seen correct without working, M0A0 otherwise |
## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\sum_{r=1}^{n} \frac{1}{4r^2-1} = \sum_{r=1}^{n}\left(\frac{1}{2r-1} - \frac{1}{2r+1}\right)$ | | |
| $= 1 - \frac{1}{3} + \frac{1}{3} - \frac{1}{5} + \frac{1}{5} - \frac{1}{7} + \ldots + \frac{1}{2n-1} - \frac{1}{2n+1} = 1 - \frac{1}{2n+1}$ | M1A1ft | M1: showing fractions with PFs, min 2 at start and 1 at end, must start at 1 and end at $n$; A1ft: identify 2 non-cancelling fractions |
| $= \frac{2n+1-1}{2n+1}$ | | |
| $\sum_{r=1}^{n} \frac{1}{4r^2-1} = \frac{n}{2n+1}$ | A1 | A1cso: correct final answer |
---\begin{enumerate}
\item (a) Express $\frac { 2 } { 4 r ^ { 2 } - 1 }$ in partial fractions.\\
(b) Hence use the method of differences to show that
\end{enumerate}
$$\sum _ { r = 1 } ^ { n } \frac { 1 } { 4 r ^ { 2 } - 1 } = \frac { n } { 2 n + 1 }$$
\hfill \mbox{\textit{Edexcel FP2 2014 Q1 [5]}}