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

Using the hint given by Almot1960 we have $\sin^4 x =\frac{3}{8} \color{red}{-\frac{1}{2} \cos(2x)} \color{blue}{ +\frac{1}{8} \cos (4x) }$ ...
\begin{eqnarray*}
\sin^4 x +\frac{1}{4} \sin^4 2x + \frac{1}{4^2} \sin^4 4x +\frac{1}{4^3} \sin^4 8x + \cdots \\
=\frac{3}{8} \color{red}{-\frac{1}{2} \cos(2x)} \color{blue}{ +\frac{1}{8} \cos (4x) } +\frac{3}{8} \frac{1}{4} \color{blue}{-\frac{1}{8} \cos(4x)} \color{red}{ +\frac{1}{32} \cos (8x) } +\frac{3}{8}\frac{1}{4^2} \color{red}{-\frac{1}{32} \cos(8x)} \color{blue}{ +\frac{1}{128} \cos (16x) } \cdots 
\end{eqnarray*}
So you were absolutely right , it is a telescoping sum & lots of the terms cancel. The terms in black are simply a geometric series
\begin{eqnarray*}
 \frac{3}{8} \left( 1+ \frac{1}{4}+\frac{1}{4^2}+ \cdots \right)= \frac{1}{2}
\end{eqnarray*}
So the sum is $ \frac{1- \cos(2x)}{2} =\sin^2 (x)$ and the answer is $\color{red}{(B)}$.
