Answer
$\Sigma_{n=0}^\infty nx^{(n-1)}$
Work Step by Step
Consider the Taylor Series for $\dfrac{1}{1-x}$ as follows:
$\dfrac{1}{1-x}=\Sigma_{n=0}^\infty x^n =1+x+x^2+....+x^n$
Now, $ (\dfrac{1}{1-x})'=[1+x+x^2+....+x^n]'$
Use formula $\dfrac{d}{dx} (x^n) =n x^{n-1}$
Then, we have $\dfrac{1}{(1-x)^2}=1+2x+3x^2+.....nx^{n-1}$
$\implies \Sigma_{n=0}^\infty nx^{(n-1)}$