Using differentiation to find the power series for $x/(1+2x)^2$ So I am not sure how to go about this problem: finding the series representation of
$$f(x) = \frac{x}{(1+2x)^2}$$
I used $1/(1+2x)$ and created the power series
$$\sum_{n=0}^\infty (-2x)^n = \frac{1}{1+2x}$$
and differentiated both sides. My derivative was
$$\sum_{n=0}^\infty (n+1)(-1)^{n+1}(2x)^n = \frac{-2}{(1+2x)^2}$$
I found that I needed to multiply my power series by $-x/2$ in order for the power series to be equivalent to $x/(1+2x)^2$. What I don't understand is why this is wrong. Can someone help point me in the right direction?

$\frac{1}{1+2x} = \sum _0^{inf} \left(-2x \right )^n$
$-\frac{2}{\left ( 1+2x \right )^2} = \sum _1^{inf}{n \times \left( -2\right )^n \times x^{n-1}}=\sum _0^{inf}{\left( n+1\right ) \times \left( -2\right )^{n+1} \times x^{n}}$
