Find $\lim_{n\to\infty}\sin^4x+\frac{1}{4}\sin^42x+.....+\frac{1}{4^n}\sin^42^nx$ $\lim_{n\to\infty}\sin^4x+\frac{1}{4}\sin^42x+.....+\frac{1}{4^n}\sin^42^nx$
I feel that this question will be solved by telescoping series but i cannot express it as telescoping series.
$\sin^4x=(\frac{1-\cos2x}{2})^2=\frac{1+\cos^22x-2\cos2x}{4}$
I am not able to solve it further.
The options are $(A)\sin^4x,(B)\sin^2x,(C)\cos^2x$(D)does not exist

Consider : $\displaystyle \frac{\sin^{4}(2^{i}x)}{4^i} + \frac{\sin^{4}(2^{i+1}x)}{4^{i+1}} = \frac{1}{4^{i}}\big(\frac{1}{8}\cos(4\cdot2^{i}x) -\frac{1}{2}\cos(2\cdot2^{i}) + \frac{3}{8} + \frac{1}{32}\cos(2\cdot2^{i+1})-\frac{1}{8}\cos(2\cdot2^{i+1}) + \frac{3}{8}\big)$
Now you could subtract two terms in neighbour parts.
