<s>[INST] Inverse Z-Transform of $(1+2/z)^{-3}$ Working through the K.A Stroud Advanced Engineering Mathematics Textbook on my own, and have really got stuck on this question with no one to help out.
The question and my working can be found here
Basically, brought it down into 3 fractions, which I could transform up until the last one $\frac{4z}{(z+2)^3}$ , as can be seen in the top corner.
If anyone knows the inverse Z-transform of $\frac{4z}{(z+2)^3}$, but not necessarily the answer to the main question it would still be really appreciated.
I tried going with the derivative rule where if $Z(x)=F(z)$ then -$zF'(z)=Z(kx)$, and ended up with a final answer of $(-2)k * (z+1)^2$ which I know was wrong. I believe it should be $(-2)k * (z+2)^2$.
Any help would be greatly appreciated as I have no access to any professors or fellow students. Thanks so much!
 [/INST] $$\frac{z}{(z+2)^3}=\frac{z^{-2}}{(1+\frac{2}{z} )^3}=z^{-2}\sum_{k=0}^\infty \binom{k+2}{k}\left(-\frac{2}{z}\right)^k=
z^{-2}\sum_{k=0}^\infty \frac{1}{2}(k+1)(k+2)\left(-\frac{2}{z}\right)^k$$
$$\frac{z}{(z+2)^3}=\sum_{k=0}^\infty (-1)^k 2^{k-1}(k+1)(k+2)\frac{1}{z^{k+2}}$$
$k=n-2$
$$\frac{z}{(z+2)^3}=\sum_{n=-2}^\infty (-1)^{n-2} 2^{n-3}(n-1)(n)\frac{1}{z^{n}}=\sum_{n=-2}^\infty \frac{a_n}{z^{n}}$$
$$\mathcal{Z}^{-1}\left(\frac{z}{(z+2)^3}\right)_{(n)}=a_n=(-1)^{n} 2^{n-3}(n-1)n \qquad\qquad n\geq 0$$
$\mathcal{Z}^{-1}$ denotes the inverse Z-transform.
$$ $$
$$\frac{1}{(1+\frac{2}{z} )^3}=\sum_{k=0}^\infty \binom{k+2}{k}\left(-\frac{2}{z}\right)^k=
\sum_{k=0}^\infty \frac{1}{2}(k+1)(k+2)\left(-\frac{2}{z}\right)^k$$
$$\frac{1}{(1+\frac{2}{z} )^3}=\sum_{k=0}^\infty (-1)^k 2^{k-1}(k+1)(k+2)\frac{1}{z^{k}}$$
$$\mathcal{Z}^{-1}\left(\frac{1}{(1+\frac{2}{z} )^3}\right)_{(n)}=(-1)^{n} 2^{n-1}(n+1)(n+2) \qquad\qquad n\geq 0$$
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