| Exam Board | CAIE |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2009 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Partial Fractions |
| Type | Repeated squared factors in denominator |
| Difficulty | Standard +0.8 This is a telescoping series question requiring verification of an algebraic identity, recognition of the telescoping pattern, and evaluation of partial sums. While the verification is straightforward algebra and the telescoping mechanism is given, students must carefully track terms through summation and take limits. It's moderately challenging for Further Maths, requiring careful algebraic manipulation but following a standard pattern once the telescoping structure is identified. |
| Spec | 4.06b Method of differences: telescoping series |
| Answer | Marks |
|---|---|
| Verifies displayed result | M1A1 |
| (i) \(S_N = \frac{1}{15} - \frac{1}{(N+3)(2N+5)}\) | M1A1A1 |
| (ii) \(S_\infty = \frac{1}{15}\) | A1 ft |
| Answer | Marks | Guidance |
|---|---|---|
| Either \((2n^2 + 11n + 15) - (2n^2 + 7n + 6)\) (oe) | A1 | in numerator |
| Or \((n+3)(2n+5) - (n+2)(2n+3)\) | A1 | in numerator |
Verifies displayed result | M1A1 |
**(i)** $S_N = \frac{1}{15} - \frac{1}{(N+3)(2N+5)}$ | M1A1A1 |
**(ii)** $S_\infty = \frac{1}{15}$ | A1 ft |
**Note:** Must see working for preliminary result
Either $(2n^2 + 11n + 15) - (2n^2 + 7n + 6)$ (oe) | A1 | in numerator
Or $(n+3)(2n+5) - (n+2)(2n+3)$ | A1 | in numerator2 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\\
(i) the sum to $N$ terms,\\
(ii) the sum to infinity.
\hfill \mbox{\textit{CAIE FP1 2009 Q2 [6]}}