Challenging +1.2 This is a Further Maths question requiring geometric series summation and de Moivre's theorem application. Part (i) is straightforward GP formula recall. Part (ii) requires substituting z = e^(iπ/10), separating real/imaginary parts, and algebraic manipulation—standard FM techniques but more steps than typical A-level. The multi-step nature and FM content place it above average difficulty.
Write down an expression in terms of \(z\) and \(N\) for the sum of the series
$$\sum_{n=1}^N 2^{-n}z^n.$$ [2]
Use de Moivre's theorem to deduce that
$$\sum_{n=1}^{10} 2^{-n}\sin\left(\frac{1}{10}n\pi\right) = \frac{1025\sin\left(\frac{1}{10}\pi\right)}{2560 - 2048\cos\left(\frac{1}{10}\pi\right)}.$$ [6]
Write down an expression in terms of $z$ and $N$ for the sum of the series
$$\sum_{n=1}^N 2^{-n}z^n.$$ [2]
Use de Moivre's theorem to deduce that
$$\sum_{n=1}^{10} 2^{-n}\sin\left(\frac{1}{10}n\pi\right) = \frac{1025\sin\left(\frac{1}{10}\pi\right)}{2560 - 2048\cos\left(\frac{1}{10}\pi\right)}.$$ [6]
\hfill \mbox{\textit{CAIE FP1 2005 Q7 [8]}}