Inverse z of (2z³+z)/((z-2)²(z-1))

X(z) = (2z³ + z)/((z − 2)²(z − 1)).

Step 1. Divide by z and use partial fractions (a double pole at 2 and a simple pole at 1):

X(z)/z = (2z² + 1)/((z − 2)²(z − 1)) = A/(z − 1) + B/(z − 2) + C/(z − 2)²

• A = (2z² + 1)/(z − 2)² at z = 1 = 3/1 = 3
• C = (2z² + 1)/(z − 1) at z = 2 = 9/1 = 9
• B: for large z, X/z ≈ 2/z, so A + B = 2, which gives B = −1

(Check at z = 0: the left side is 1/((4)(−1)) = −1/4 and the right side is −3 + 1/2 + 9/4 = −1/4. ✓)

Step 2. Multiply by z:

X(z) = 3z/(z − 1) − z/(z − 2) + 9z/(z − 2)²

Step 3. Invert with the standard pairs:

• Z⁻¹[z/(z − 1)] = 1
• Z⁻¹[z/(z − 2)] = 2ᵏ
• Z⁻¹[z/(z − a)²] = k aᵏ⁻¹, so Z⁻¹[z/(z − 2)²] = k 2ᵏ⁻¹

Answer:

x(k) = 3 − 2ᵏ + 9k · 2ᵏ⁻¹, k ≥ 0

k    : 0   1   2    3    4
x(k) : 2  10  35  103  275

(Checked by long division: X(z) = 2 + 10z⁻¹ + 35z⁻² + 103z⁻³ + …)