What is the value of the infinite sum ∑n=1∞sin(nθ)2n\sum_{n=1}^{\infty} \frac{\sin(n\theta)}{2^n}∑n=1∞2nsin(nθ)?
2sinθ5−4cosθ\frac{2\sin\theta}{5 - 4\cos\theta}5−4cosθ2sinθ
sinθ5−4cosθ\frac{\sin\theta}{5 - 4\cos\theta}5−4cosθsinθ
cosθ5−4sinθ\frac{\cos\theta}{5 - 4\sin\theta}5−4sinθcosθ
15−4cosθ\frac{1}{5 - 4\cos\theta}5−4cosθ1