Z-transform of k² a^(k−1)

Use the known pair $\mathcal{Z}[a^k] = \frac{z}{z-a}$ and the multiplication by $k$ property $\mathcal{Z}[k\,x(k)] = -z\frac{d}{dz}X(z)$.

Step 1: $\mathcal{Z}[k a^{k-1}]$

$$\mathcal{Z}[k a^k] = -z\frac{d}{dz}\left(\frac{z}{z-a}\right) = -z\cdot\frac{(z-a) - z}{(z-a)^2} = \frac{az}{(z-a)^2}$$

so $\mathcal{Z}[k a^{k-1}] = \frac{1}{a}\cdot\frac{az}{(z-a)^2} = \frac{z}{(z-a)^2}$.

Step 2: $\mathcal{Z}[k^2 a^{k-1}]$

$$\begin{aligned} \mathcal{Z}[k\cdot k a^{k-1}] &= -z\frac{d}{dz}\left[\frac{z}{(z-a)^2}\right] \\ &= -z\cdot\frac{(z-a)^2 - 2z(z-a)}{(z-a)^4} \\ &= -z\cdot\frac{(z-a) - 2z}{(z-a)^3} = \frac{z(z+a)}{(z-a)^3} \end{aligned}$$

Check

Series of $\frac{z(z+a)}{(z-a)^3} = z^{-1} + 4a z^{-2} + 9a^2 z^{-3} + \cdots$, i.e. coefficients $k^2a^{k-1}$ ✔.

Answer: $\mathcal{Z}[k^2a^{k-1}] = \dfrac{z(z+a)}{(z-a)^3} = \dfrac{z^{-1}(1+az^{-1})}{(1-az^{-1})^3}$, $|z| > |a|$.