Inverse z transform - help
You know the geometric series

$$\sum_{n=0}^{\infty}r^n=\frac{1}{1-r},\quad |r|&lt;1\tag{1}$$

Due to convergence of (1) (for $|r|&lt;1$) 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

$$\sum_{n=0}^{\infty}nr^{n-1}=\sum_{n=1}^{\infty}nr^{n-1}=\sum_{n=0}^{\infty}(n+1)r^n=\frac{1}{(1-r)^2},\quad |r|&lt;1\tag{2}$$

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 $|z|&gt;a=2$.