I have to $\,Z$-transform the sequence
$$a(n)=\begin{cases} (n-2)^2\quad\text{ if }\,n\geq 2\\0\qquad\quad\text{ elsewhere }\end{cases}$$
I know that I have to use the time-shift rule, since $(n-2)^2$ is a shifted version of $n^2$ by two-steps.
So the shift-rule says $$y(n-k)\Theta(n-k)=z^{-k}Y(z)$$
Clearly, the first term is $z^{-2}$, and the relevant rule of transform for the other part is $$a^n\rightarrow \frac{z}{z-a}$$ But I can't find the correct way to use it.
Any ideas?
Thanks
$\begin{align}\displaystyle Z[a(n)]&=\sum_{n=2}^{\infty}(n-2)^2z^{-n}\!\!\!\underset{\overbrace{\text{by letting }m=n-2}}{=}\,\sum_{m=0}^{\infty}m^2z^{-m-2}=\\ &=\sum_{m=0}^{\infty}\big[m(m+1)-m\big]z^{-m-2}=\\ &=\sum_{m=0}^{\infty}m(m+1)z^{-m-2}-\sum_{m=0}^{\infty}mz^{-m-2}=\\ &=\sum_{m=0}^{\infty}(-m)(-m-1)z^{-m-2}+z^{-1}\!\sum_{m=0}^{\infty}(-m)z^{-m-1}=\\ &=\sum_{m=0}^{\infty}\dfrac d{dz}\left[(-m)z^{-m-1}\right]+z^{-1}\!\sum_{m=0}^{\infty}\frac d{dz}\left(z^{-m}\right)=\\ &=\sum_{m=0}^{\infty}\dfrac{d^2}{dz^2}\left(z^{-m}\right)+z^{-1}\!\sum_{m=0}^{\infty}\frac d{dz}\left(z^{-m}\right)=\\ &=\dfrac{d^2}{dz^2}\left(\sum_{m=0}^{\infty}z^{-m}\!\right)+z^{-1}\frac d{dz}\left(\sum_{m=0}^{\infty}z^{-m}\!\right)=\\ &=\dfrac{d^2}{dz^2}\left(\!\frac1{1-z^{-1}}\!\right)+z^{-1}\frac d{dz}\left(\!\frac1{1-z^{-1}}\!\right)=\\ &=\dfrac{d^2}{dz^2}\left(\!\frac z{z-1}\!\right)+\frac1z\!\cdot\!\frac d{dz}\left(\!\frac z{z-1}\!\right)=\\ &=\dfrac d{dz}\left[\frac{-1}{(z-1)^2}\right]+\frac1z\!\cdot\!\frac{-1}{(z-1)^2} =\\ &=\frac2{(z-1)^3}-\frac1{z(z-1)^2}=\\ &=\frac{2z-(z-1)}{z(z-1)^3}=\\ &=\dfrac{z+1}{z(z-1)^3}\;.\end{align}$