Derivative of Geometric Sequence/Corollary

Theorem
Let $x \in \R: \left|{x}\right| < 1$.

Then:
 * $\displaystyle \sum_{n \mathop \ge 1} n \left({n+1}\right) x^{n-1} = \frac 2 {\left({1-x}\right)^3}$

Proof
We have from Power Rule for Derivatives that:
 * $\displaystyle \frac {\mathrm d} {\mathrm d x} \sum_{n \mathop \ge 1} \left({n+1}\right) x^n = \sum_{n \mathop \ge 1} n \left({n+1}\right) x^{n-1}$

But from Sum of Infinite Geometric Progression:
 * $\displaystyle \sum_{n \mathop \ge 1} \left({n+1}\right) x^n = \frac 1 {1-x}$

The result follows by Power Rule for Derivatives and the Chain Rule applied to $\dfrac 1 {\left({1-x}\right)^2}$.