# $\displaystyle\sum_{k=0}^n \frac{\cos(k x)}{\cos^kx} = ?$

Prove that (not use induction) $\displaystyle\sum_{k=0}^n \frac{\cos(k x)}{\cos^kx} = \frac{1+(-1)^n}{2\cos^nx} + \dfrac{2\sin\big(\lfloor\frac{n+1}{2}\rfloor x\big) \cos\big(\lfloor\frac{n+2}{2}\rfloor x\big)} {\sin x\cos^n x} \qquad\qquad (\frac{2x}{\pi}\not\in \mathbb Z)$

• Why should we not use induction? – GEdgar Sep 21 '13 at 13:10

Euler Formula says $e^{iy}=\cos y+\sin y,$
$\sum_{k=0}^n \frac{\cos(k x)}{\cos^kx} =$Re $\left(\sum_{k=0}^n \left(\frac{e^{ix}}{\cos x}\right)^k\right)$
Now, $\sum_{k=0}^n \left(\frac{e^{ix}}{\cos x}\right)^k$ is a Geometric Series
Now, deal the even & the odd cases of $n$ separately