| Exam Board | CAIE |
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2005 |
| Session | November |
| Topic | Reduction Formulae |
6 Let
$$I _ { n } = \int _ { 0 } ^ { 1 } \left( 1 + x ^ { 2 } \right) ^ { - n } \mathrm {~d} x$$
where \(n \geqslant 1\). By considering \(\frac { \mathrm { d } } { \mathrm { d } x } \left( x \left( 1 + x ^ { 2 } \right) ^ { - n } \right)\), or otherwise, prove that
$$2 n I _ { n + 1 } = ( 2 n - 1 ) I _ { n } + 2 ^ { - n }$$
Deduce that \(I _ { 3 } = \frac { 3 } { 32 } \pi + \frac { 1 } { 4 }\).
\(7 \quad\) Write down an expression in terms of \(z\) and \(N\) for the sum of the series
$$\sum _ { n = 1 } ^ { N } 2 ^ { - n } z ^ { n }$$
Use de Moivre's theorem to deduce that
$$\sum _ { n = 1 } ^ { 10 } 2 ^ { - n } \sin \left( \frac { 1 } { 10 } n \pi \right) = \frac { 1025 \sin \left( \frac { 1 } { 10 } \pi \right) } { 2560 - 2048 \cos \left( \frac { 1 } { 10 } \pi \right) }$$