sequences and series - How do I find the sum of $sumlimits_{k=1}^infty{frac{k}{2^{k+1}}}=1$?

As shown in the title, how do I find the sum of:

$$\sum\limits_{k=1}^\infty{\frac{k}{2^{k+1}}}=1$$

Answer

HINT:

Note that for $|x|<1$, $f(x)=\sum_{k=1}^{\infty}x^{k}=\frac{x}{1-x}$ implies that

$$x^2f'(x) = \sum_{k=1}^{\infty}kx^{k+1}$$

Then, let $x=1/2$