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?
First, how I would approach this:
You'll want to start the geometric series, $|x| &lt; 1$,
$$\sum_{n=0}^\infty x^n = \frac{1}{1-x}$$
Differentiate both sides with respect to $x$, and then multiply both sides by $x$. You will obtain
$$\sum_{n=0}^\infty nx^n = \frac{x}{(1-x)^2}$$
Replace $x$ with $-2x$:
$$\sum_{n=0}^\infty n(-2x)^n = \frac{-2x}{(1+2x)^2}$$
Multiply both sides by $-1/2$:
$$- \frac 1 2 \sum_{n=0}^\infty n(-2x)^n = \frac{x}{(1+2x)^2}$$
You can bring the $-1/2$ into the summation since $(-2x)^n = (-1)^n 2^n x^n$, causing some simplification:
$$\sum_{n=0}^\infty n(-1)^{n+1} 2^{n-1} x^n= \frac{x}{(1+2x)^2}$$
Of course, $(-1)^{n+1} = (-1)^{n-1}$, so for a slightly cleaner-looking final answer, you can use that fact to recombine the $-1$ and $2$:
$$\sum_{n=0}^\infty n(-2)^{n-1} x^n= \frac{x}{(1+2x)^2}$$

Your work, and where your error seems to arise:
You begin by claiming, via the geometric series,
$$\sum_{n=0}^\infty (-2x)^n = \frac{1}{1+2x}$$
Differentiating with respect to $x$ on both sides, you should get
$$\sum_{n=0}^\infty (-2)n(-2x)^{n-1} = \frac{-2}{(1+2x)^2}$$
The summation simplifies a bit to
$$\sum_{n=0}^\infty (-2)n(-2x)^{n-1} = \sum_{n=0}^\infty n(-2)^n x^{n-1}$$
which doesn't at all seem like the result you claimed of
$$\sum_{n=0}^\infty (n+1)(-1)^{n+1} (2x)^n$$
So I assume a calculation error arises here.
$\frac{1}{1+2x} = \sum _0^{inf} \left(-2x \right )^n$
$-\frac{2}{\left ( 1+2x \right )^2} = \sum _1^{inf}{n \times \left( -2\right )^n \times x^{n-1}}=\sum _0^{inf}{\left( n+1\right ) \times \left( -2\right )^{n+1} \times x^{n}}$