Inverse z of (z+2)/((z-2)(z-3))

Partial fractions of X(z)/z.

X(z)/z = (z + 2)/[z (z − 2)(z − 3)] = A/z + B/(z − 2) + C/(z − 3)

A = (0 + 2)/[(0 − 2)(0 − 3)] = 2/6  = 1/3
B = (2 + 2)/[2 (2 − 3)]      = 4/(−2) = −2
C = (3 + 2)/[3 (3 − 2)]      = 5/3

X(z) = 1/3 − 2 z/(z − 2) + (5/3) z/(z − 3)

Inverse: the constant 1/3 gives (1/3)δ(k); z/(z − a) gives aᵏ.

x(k) = (1/3) δ(k) − 2·2ᵏ + (5/3)·3ᵏ

For k ≥ 1 this simplifies to the neat form

Answer: x(k) = 5·3^{k−1} − 4·2^{k−1} (k ≥ 1), and x(0) = 0.

Check: expanding X(z) in powers of z⁻¹ gives x(0..4) = 0, 1, 7, 29, 103, and the formula gives 5·3^{k−1} − 4·2^{k−1} = 1, 7, 29, 103 for k = 1..4 ✓.