How to prove $\sum_{k=0}^{\infty}k^2x^{k} = \frac{x(1+x)}{(1-x)^3}\text{, }|x| < 1$?

Question

How do I prove that the summation $$\sum_{k=0}^{\infty}k^2x^{k} = \dfrac{x(1+x)}{(1-x)^3}\text{, }|x| < 1\text{?}$$

Answer 1

$$\sum_{k=0}^{\infty}x^k = \frac{1}{1-x}$$

Taking derivatives both sides with respect to $x$, the first and the second derivatives yield respectively:

$$\sum_{k=1}^{\infty}kx^{k-1} = \frac{1}{(1 - x)^2}$$

$$\sum_{k=2}^{\infty}k(k-1)x^{k-2} = \frac{2}{(1-x)^3}$$

Now:

$$\sum_{k=0}^{\infty}k^2x^k = \sum_{k=1}^{\infty}k^2x^k = \sum_{k=1}^{\infty}(k^2 - k + k)x^k = \sum_{k=1}^{\infty}k(k-1)x^k + \sum_{k=1}^{\infty}kx^k = \sum_{k=2}^{\infty}k(k-1)x^k + \sum_{k=1}^{\infty}kx^k = x^2\sum_{k=2}^{\infty}k(k-1)x^{k-2} + x\sum_{k=1}^{\infty}kx^{k-1} = \frac{2x^2}{(1-x)^3} + \frac{x}{(1 - x)^2} = \frac{2x^2 + x(1 - x)}{(1-x)^3} = \frac{x^2 + x}{(1-x)^3} = \frac{x(1 + x)}{(1-x)^3}$$

Answer 2

HINT:

For $|x|<1,$ $$\sum_{k=0}^\infty x^k=\dfrac1{1-x}$$

Differentiate both sides wrt $x$

Multiply by $x$

Differentiate both sides wrt $x$

and so on

Answer 3

Write $$ \frac{x(1+x)}{(1-x)^3}=\frac{x^2-2x+1+3x-1}{(1-x)^3}=\frac{1}{1-x}+3x\frac{1}{(1-x)^3}-\frac{1}{(1-x)^3}. $$ Let's begin with $$ \frac{1}{1-x}=\sum_{k=0}^\infty x^k. $$ Differentiate $2$ times, which gives $$ \frac{2}{(1-x)^3}=\sum_{k=2}^\infty k(k-1)x^{k-2}=\sum_{k=0}^\infty(k+2)(k+1)x^k. $$ Thus, your RHS equals to $$ S\equiv\sum_{k=0}^\infty x^k+\frac{3x}{2}\sum_{k=0}^\infty(k+2)(k+1)x^k-\frac{1}{2}\sum_{k=0}^\infty(k+2)(k+1)x^k. $$ The coefficient corresponding to $x^0$ is $1-\frac{1}{2}2\times 1=0=0^2$ (i.e. the middle expression above does not contribute.). The coefficient corresponding to $x^k$ for $k\geq 1$ is $$ 1+\frac{3}{2}(k+1)k-\frac{1}{2}(k+2)(k+1)=k^2. $$ The claim follows.