2024/003) Find $\sum_{n=1}^{\infty} \frac{n^2}{10^n}$

Here let $frac{1}{10} = x$

 We have for $|x|  \lt 1$

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

Differentiating both sides wrt x we get

$\sum_{n=0}^{\infty} nx^{n-1}= \frac{1}{(1-x)^2}$

Multiplying both sides by x we get 

 $\sum_{n=0}^{\infty} nx^n= \frac{x}{(1-x)^2}$

Differenting both sides wrt x we get

$\sum_{n=0}^{\infty} n^2x^{n-1}= \frac{d}{dx} \frac{x}{(1-x)^2}$

Usimg $\frac{d} ({dx} \frac{u}{v})= \frac{v\frac{du}{dx} - u \frac{dv}{dx}}{v^2}$

We get 

 $\sum_{n=0}^{\infty} n^2x^{n-10} =\frac{(1-x)^2 . 1 - x . (x-2)}{(1-x)^4} = \frac{x +1}{(1-x)^3}$

Multiply by x on both sides we get $\sum_{n=0}^{\infty} n^2x^n = \frac{x^2 +x}{(1-x)^3}$

Putting value of $x = \frac{1}{10}$ we get the result