Find a power series for $\frac{z^2}{(4-z)^2}$.

Question

Find a power series for $$\frac{z^2}{(4-z)^2}$$ What I did is: Since $$\sum^{\infty}_{k=0}z^k=\frac{1}{1-z}$$ Take the derivative, so $$\sum^{\infty}_{k=1}kz^{k-1}=\frac{1}{(1-z)^2}$$ So $$\frac{z^2}{(4-z)^2}=\frac{z^2}{4^2(1-\frac{z}{4})^2}=\frac{z^2}{4^2}\sum^{\infty}_{k=1}k\left(\frac{z}{4}\right)^{k-1}=\sum^{\infty}_{k=1}k\left(\frac{z}{4}\right)^{k+1}$$ However, the solution is $$\sum^{\infty}_{k=0}\frac{k+1}{2\cdot 4^{k}}z^{k+2}$$ Thanks for any comments~

Answer 1

For $|z|<4$ we have:$$\frac{1}{4-z}=\sum_{n=0}^{\infty}\frac{z^n}{4^{n+1}}\to \frac{1}{(4-z)^2}=\sum_{n=0}^{\infty}\frac{(n+1)z^n}{4^{n+2}}\to\frac{z^2}{(4-z)^2}=\sum_{n=0}^{\infty}\frac{(n+1)z^{n+2}}{4^{n+2}}$$ and for $|z|>4$ similarly: $$\frac{z^2}{(4-z)^2}=\sum_{n=0}^{\infty}\frac{(n+1)4^n}{z^{n}}$$