Z-transform of 9k·2^(k−1) − 2^k + 3

Use linearity and three standard pairs.

Standard pairs used

The last pair comes from $\mathcal{Z}[k\,x(k)] = -z\dfrac{d}{dz}X(z)$:

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

Dividing by $a$ gives $\mathcal{Z}[k a^{k-1}] = \dfrac{z}{(z-a)^2}$.

Transform each term ($a = 2$)

$$\begin{aligned} \mathcal{Z}[9k\,2^{k-1}] &= \frac{9z}{(z-2)^2} \\ \mathcal{Z}[-2^k] &= -\frac{z}{z-2} \\ \mathcal{Z}[3] &= \frac{3z}{z-1} \end{aligned}$$ $$X(z) = \frac{9z}{(z-2)^2} - \frac{z}{z-2} + \frac{3z}{z-1}$$

Combine over a common denominator $(z-2)^2(z-1)$

$$\begin{aligned} \text{Numerator} &= 9z(z-1) - z(z-2)(z-1) + 3z(z-2)^2 \\ &= z\left[9z - 9 - (z^2 - 3z + 2) + 3(z^2 - 4z + 4)\right] \\ &= z\left[2z^2 + 0\cdot z + 1\right] \end{aligned}$$ $$X(z) = \frac{z(2z^2 + 1)}{(z-2)^2(z-1)}$$

Check

From the formula: $x(0) = 0 - 1 + 3 = 2$, $x(1) = 9 - 2 + 3 = 10$, $x(2) = 36 - 4 + 3 = 35$, $x(3) = 108 - 8 + 3 = 103$.

Long division of $X(z)$ gives $2 + 10z^{-1} + 35z^{-2} + 103z^{-3} + \dots$, which matches.

Answer: $X(z) = \dfrac{9z}{(z-2)^2} - \dfrac{z}{z-2} + \dfrac{3z}{z-1} = \dfrac{z(2z^2+1)}{(z-2)^2(z-1)}$, ROC $|z| > 2$.