Finding maclaurin series

I'm trying to find the MacLaurin series for the following:

$$\frac{1}{(1+2z)^2}$$

What I'm trying to do is to take the integral of that, find the Maclaurin series then derivate it, but I'm not sure that's valid for this kind of series.

Here's my attempt:

$$\int{f(z)} =\int{\frac{1}{(1+2z)^2}dz} = - \frac{1}{2+4z}$$

then I try to expand it:

$$\frac{1}{2+4z} = \frac{1}{1+(1+4z)}$$

taking $r =(-1)(1+4z)$

$$f(z) = \frac{d}{dz}(- \frac{1}{2+4z}) = -\frac{d}{dz}(\displaystyle\sum\limits_{k=0}^{+\infty} (-1)^k(1+4z)^k)$$

And derivating

I get the result

$$f(z) = \sum\limits_{k=0}^{+\infty} (-1)^{k+1}k(1+4z)^{k-1}$$

Doesn't look right, and I'm not really sure if I can do this (integrate then derivate) like I do with Laurent Series. Not homework but part of my exercises I'm using to study for an exam.

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+1 for showing your work! – Jyrki Lahtonen Jun 23 '11 at 14:31

You have the right idea. Instead of $$\frac{1}{2+4z}=\frac{1}{1+(1+4z)}$$ try $$\frac{1}{2+4z}=\frac12\cdot\frac{1}{1+2z},$$ and write that as a geometric series as you did in your first attempt.
Thank you, using your sugestion i've obtained the correct series, $\displaystyle\sum\limits_{k=0}^{+\infty}(-1)^k2^k(k+1)z^k$ – GriffinHeart Jun 23 '11 at 15:00