How does the answer get k(k-1) from the geometric series?

Differentiate sum w^k twice and multiply by w^2

Differentiate once and multiply by w

How is the result verified?

Z[k^2] - Z[k] = z(z+1)/(z-1)^3 - z/(z-1)^2 = 2z/(z-1)^3

Short answer

Using Σ k(k-1) w^k = 2w^2/(1-w)^3 with w=1/z yields Z[n(n-1)] = 2z/(z-1)^3, valid for |z|>1.

Find the z-transform of n(n - 1); n = k.

Meaning. The sequence is x(k) = k(k − 1), k ≥ 0 (n = k). Its z-transform is X(z) = Σ k(k − 1) z⁻ᵏ.

Method (geometric series). For |w| < 1:

Σ_{k=0}^{∞} wᵏ = 1/(1 − w)

Differentiate twice with respect to w and multiply by w²:

Σ k(k − 1) wᵏ = 2w² / (1 − w)³

Put w = 1/z:

X(z) = 2 z⁻² / (1 − 1/z)³
     = 2 z⁻² · z³ / (z − 1)³
     = 2z / (z − 1)³ ,   |z| > 1

Check by the standard results. Z[k²] = z(z + 1)/(z − 1)³ and Z[k] = z/(z − 1)². So

Z[k² − k] = [z(z + 1) − z(z − 1)] / (z − 1)³ = 2z/(z − 1)³ ✓

Answer: Z[n(n − 1)] = 2z/(z − 1)³.

✓ Answer checked by Claude Opus 5.5 (high effort)