Expand $\frac{1}{z^2}$ as a series on $|z - 2| < 2$ I am asked to show that 
$$\frac{1}{z^2} = \frac{1}{4} + \frac{1}{4}\sum_{n = 1}^{\infty}(-1)^{n+1}(n+1)\left(\frac{z-2}{2}\right)^n$$
on $|z - 2| < 2$.
My plan is to consider a series expansion of $\frac{-1}{z}$ in $|z - 2| < 2$ and differentiate this series term by term to yield the desired series (This is valid from theorems we have proved in class).
Indeed,
$$\frac{-1}{z} = \frac{-1}{2-2+z} = \frac{-1}{2}\cdot\frac{1}{1+\left(\frac{z-2}{2}\right)}$$
$$ = \frac{-1}{2}\sum_{n = 0}^{\infty}(-1)^n\left(\frac{z-2}{2}\right)^n $$
$$ = \frac{1}{2}\sum_{n = 0}^{\infty}(-1)^{n+1}\left(\frac{z-2}{2}\right)^n $$
Differentiating term by term gives:
$$ \frac{1}{z^2} = \frac{1}{2}\sum_{n = 1}^{\infty}(-1)^{n+1}n\left(\frac{z-2}{2}\right)^{n-1}\cdot\frac{1}{2} $$
$$ = \frac{1}{4}\sum_{n = 1}^{\infty}(-1)^{n+1}n\left(\frac{z-2}{2}\right)^{n-1}$$
Now, if we let $k = n - 1$, we can re-index the series as:
$$ = \frac{1}{4}\sum_{k = 0}^{\infty}(-1)^{k+2}(k+1)\left(\frac{z-2}{2}\right)^{k}$$
$$ = \frac{1}{4} + \frac{1}{4}\sum_{k = 1}^{\infty}(-1)^{k}(k+1)\left(\frac{z-2}{2}\right)^{k}$$
Unfortunately, the power of $(-1)$ present in my series does not match what the prompt reads. I am off by one. Did I make a mistake or is the prompt incorrect?

Perhaps a simpler approach would be to write
$$
{1 \over z^2} = {1 \over 1 - \left(1 - z^2\right)}.
$$
Letting $a = 1 - z^2$, we see that the above expression has (at least formally) the geometric series expansion:
$$
{1 \over 1 - \left(1 - z^2\right)} = {1 \over 1 - a} = \sum_{n \geq 0} a^{n}.
$$
Now expand each power
$$
a^{n} = \left(1 - z^2\right)^n
$$
using the Binomial Theorem.
