sum : $\large \sum_{k=0}^{n}\frac{\cos(kx)}{2^k}$

Question

I am trying to find a shorter solution to find the following sum:

$\large \sum_{k=0}^{n}\frac{\cos(kx)}{2^k}$

Actually I converted the whole sum : $\large \sum_{k=0}^{n}\frac{\cos(kx)}{2^k}$ = Rel $\large \sum_{k=0}^{n}(\frac{e^{ix}}{2})^k$.

I got a too long solution. Any shorter way to solve it.