Proving $1+2CZ+3C^2Z^2+....=1/(1-CZ)^2$, considering $\sum\limits_{i=1}^{\infty}c^iZ^i=(1-CZ)(1+2CZ+3C^2Z^2+....)$

Question

I'm told that we can prove this common identity for solving generating functions: $1+2CZ+3C^2Z^2+....=1/(1-CZ)^2$

Using only the property

$\sum\limits_{i=1}^{\infty}c^iZ^i=(1-CZ)(1+2CZ+3C^2Z^2+....)$

All I can think of is to move the $(1-CZ)$ to the same side as the summation and try to show the summation is equal to $1/(1-CZ)$, but it doesn't seem to work.

Answer 1

\begin{align} (1-CZ)\sum_{k\ge 1} k(CZ)^{k-1} &= \sum_{k\ge 1} k(CZ)^{k-1} - \sum_{k\ge 1} k(CZ)^{k} \\&= \sum_{k\ge 1} k(CZ)^{k-1} - \sum_{k\ge 2} (k-1)(CZ)^{k-1} \\&= 1 + \sum_{k\ge 2} (CZ)^{k-1} \\&= \sum_{k\ge 0} (CZ)^{k} = \\ (1-CZ)^2\sum_{k\ge 1} k(CZ)^{k-1} &= (1-CZ)\sum_{k\ge 0} (CZ)^{k} \\ &= \sum_{k\ge 0} (CZ)^{k} - \sum_{k\ge 0} (CZ)^{k+1} = 1 \end{align} using nothing but elementary sum manipulations.