| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/1 (Pre-U Further Mathematics Paper 1) |
| Year | 2010 |
| Session | June |
| Marks | 5 |
| Topic | Sequences and series, recurrence and convergence |
| Type | Partial fractions then method of differences |
| Difficulty | Standard +0.3 This is a standard method of differences question requiring partial fractions decomposition of 1/(4r²-1) = 1/((2r-1)(2r+1)), leading to a telescoping series. While it requires multiple steps (partial fractions, recognizing the telescoping pattern, finding the finite sum, then taking the limit), these are well-practiced techniques at Further Maths level with no novel insight required. Slightly easier than average due to its routine nature. |
| Spec | 4.06b Method of differences: telescoping series |
$\frac{1}{4r^2-1} \equiv \frac{\frac{1}{2}}{2r-1} - \frac{\frac{1}{2}}{2r+1}$ **B1**
$\sum_{r=1}^{n} \frac{1}{4r^2-1} = \frac{1}{2}\sum_{r=1}^{n}\frac{1}{2r-1} - \frac{1}{2}\sum_{r=1}^{n}\frac{1}{2r+1}$ **M1** Use of the Diff. Method
$= \frac{1}{2}\left\{1 + \frac{1}{3} + \frac{1}{5} + \cdots + \frac{1}{2n-1}\right\} - \frac{1}{2}\left\{\frac{1}{3} + \frac{1}{5} + \cdots + \frac{1}{2n-1} + \frac{1}{2n+1}\right\}$ **M1** Cancelling most terms
$= \frac{1}{2}\left\{1 - \frac{1}{2n+1}\right\}$ or $\frac{n}{2n+1}$ **A1**
$S_\infty = \frac{1}{2}$ **B1 ft**
**[5]**2 Use the method of differences to express $\sum _ { r = 1 } ^ { n } \frac { 1 } { 4 r ^ { 2 } - 1 }$ in terms of $n$, and hence deduce the sum of the infinite series
$$\frac { 1 } { 3 } + \frac { 1 } { 15 } + \frac { 1 } { 35 } + \ldots + \frac { 1 } { 4 n ^ { 2 } - 1 } + \ldots$$
\hfill \mbox{\textit{Pre-U Pre-U 9795/1 2010 Q2 [5]}}