How can I find inverse z transform of $$X(z)=\frac{5}{(z-2)^{2}}$$ ?
It is known that x[n] is causal.
EDIT: Here is what I have done. Since signal x(n) is causal, convergence of z transform of that signal will be outside of circle with radius r:
We have in bracket sum which represents z transform of signal:
But I don't know what to do next.
You know the geometric series
$$\sum_{n=0}^{\infty}r^n=\frac{1}{1-r},\quad |r|<1\tag{1}$$
Due to convergence of (1) (for $|r|<1$) you can take the derivative with respect to $r$ by taking the element-wise derivatives of the left-hand side. Equating the derivatives of both sides of (1) gives
$$\sum_{n=0}^{\infty}nr^{n-1}=\sum_{n=1}^{\infty}nr^{n-1}=\sum_{n=0}^{\infty}(n+1)r^n=\frac{1}{(1-r)^2},\quad |r|<1\tag{2}$$
With $r=az^{-1}$ you get the following $\mathcal{Z}$-transform relation:
$$(n+1)a^nu[n]\Longleftrightarrow \frac{1}{(1-az^{-1})^2}=\frac{z^2}{(z-a)^2}\tag{3}$$
where $u[n]$ is the unit step sequence. With the time-shifting property of the $\mathcal{Z}$-transform (see here) you get the following pair:
$$(n-1)a^{n-2}u[n-2]\Longleftrightarrow \frac{1}{(z-a)^2}\tag{4}$$
With (4) you immediately get
$$\frac{5}{(z-2)^2}\Longleftrightarrow 5(n-1)2^{n-2}u[n-2]$$
Of course, all of this is only valid inside the region of convergence, i.e. for $|z|>a=2$.
note, from wikipedia
$$ \mathcal{Z}\left\{ n a^n u[n] \right\} = \frac{az^{-1} }{ (1-a z^{-1})^2 } $$
see if you can make your problem fit that form. (gonna get a funky result because $a=2$.)
