| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/1 (Pre-U Further Mathematics Paper 1) |
| Session | Specimen |
| Marks | 7 |
| Topic | Sequences and series, recurrence and convergence |
| Type | Method of differences with given identity |
| Difficulty | Standard +0.3 This is a straightforward telescoping series question where the identity is given. Part (i) is routine algebraic verification, part (ii) applies the standard method of differences with most terms canceling, and part (iii) requires taking a simple limit. While it involves multiple steps and partial fractions concepts, no novel insight is needed—it's a textbook application of a well-known technique, making it slightly easier than average. |
| Spec | 4.06b Method of differences: telescoping series |
**(i)** Shows displayed result M1
AG requires intermediate line of working A1 **[2]**
**(ii)** $\left(\frac{1}{3\times5} - \frac{1}{4\times7}\right) + \left(\frac{1}{4\times7} - \frac{1}{5\times9}\right) + \cdots + \left(\frac{1}{(N+2)(2N+3)} - \frac{1}{(N+3)(2N+5)}\right)$ M1A1
$S_N = \frac{1}{15} - \frac{1}{(N+3)(2N+5)}$ (or equivalent) A1A1 **[4]**
**(iii)** $S_\infty = \frac{1}{15}$ follow through from **(ii)** A1ft **[1]**2 (i) Verify that, for all positive values of $n$,
$$\frac { 1 } { ( n + 2 ) ( 2 n + 3 ) } - \frac { 1 } { ( n + 3 ) ( 2 n + 5 ) } = \frac { 4 n + 9 } { ( n + 2 ) ( n + 3 ) ( 2 n + 3 ) ( 2 n + 5 ) }$$
For the series
$$\sum _ { n = 1 } ^ { N } \frac { 4 n + 9 } { ( n + 2 ) ( n + 3 ) ( 2 n + 3 ) ( 2 n + 5 ) }$$
find\\
(ii) the sum to $N$ terms,\\
(iii) the sum to infinity.
\hfill \mbox{\textit{Pre-U Pre-U 9795/1 Q2 [7]}}