Verification of Laurent series for the function $f(z)=\frac{2}{(z+2)^2}-\frac{5}{z-4}$ Suppose that $$f(z)=\frac{2}{(z+2)^2}-\frac{5}{z-4}.$$ Find the Laurent series for $f$ in powers of $z-2$ that converges when $z=1$.
This is how I approached the problem:
\begin{align}
f(z)&=-2\ \frac{d}{dz}\left(\frac{1}{z+2}\right)-\frac{5}{(z-2)-2} \\
&=-2\ \frac{d}{dz}\left(\frac{1}{4}\left(\frac{1}{1+\frac{(z-2)}{4}}\right)\right)+\frac{5}{2}\left(\frac{1}{1-\frac{(z-2)}{2}}\right) \\
&=-2\ \frac{d}{dz}\left(\frac{1}{4}\sum_{n=0}^{\infty} (-1)^n\left(\frac{z-2}{4}\right)^n\right)+\frac{5}{2}\left(\sum_{n=0}^{\infty} \left(\frac{z-2}{2}\right)^n\right) \\
&=-2\sum_{n=1}^{\infty} (-1)^n\frac{n(z-2)^{n-1}}{4^{n+1}}+5\sum_{n=0}^{\infty} \frac{(z-2)}{2^{n+1}}^n.
\end{align}
I believe this is the right answer, yet the answer provided (using a different method) states the answer is $$f(z)=\sum_{n=0}^{\infty}c_n(z-2)^n, \ \ c_n=(-1)^n\frac{2(n+1)}{4^{n+3}}+\frac{5}{2^{n+1}}.$$

After the last $=$ sign, what you got was:$$-2\sum_{n=1}^\infty(-1)^n\frac{n(z-2)^{n-1}}{4^{n+1}}.$$Since you have that $-2$ outside the sum, this is equal to$$\sum_{n=1}^\infty(-1)^{n-1}\frac{2n(z-2)^{n-1}}{4^{n+1}}=\sum_{n=0}^\infty(-1)^n\frac{2(n+1)(z-2)^n}{4^{n+2}}.$$So, the provided answer is not correct.
