Find the value of the infinite sum ∑n=1∞cos(nθ)2n\sum_{n=1}^{\infty} \frac{\cos(n\theta)}{2^n}∑n=1∞2ncos(nθ).
2cosθ−15−4cosθ\frac{2\cos\theta - 1}{5 - 4\cos\theta}5−4cosθ2cosθ−1
cosθ−0.51.25−cosθ\frac{\cos\theta - 0.5}{1.25 - \cos\theta}1.25−cosθcosθ−0.5
2cosθ−14−4cosθ\frac{2\cos\theta - 1}{4 - 4\cos\theta}4−4cosθ2cosθ−1
12cosθ\frac{1}{2\cos\theta}2cosθ1