Answer
$\Sigma_{n=0}^\infty (n+1)(n+2)x^{(n)}$
Work Step by Step
Consider the Taylor Series for $\dfrac{1}{(1-x)^2}$ as follows:
$\dfrac{1}{(1-x)^2}=1+2x+3x^2+.....nx^{n-1}$
Then, we have $ \dfrac{d}{dx}[\dfrac{1}{(1-x)^2}]=\dfrac{d}{dx}[1+2x+3x^2+.....nx^{n-1}]$
or, $=2+6x+12x^2+....n(n+1)x^{n-1}$
This implies that $\dfrac{2}{(1-x)^3}=\Sigma_{n=0}^\infty (n+1)(n+2)x^{(n)}$