Answer
$\Sigma_{n=0}^\infty (n+1)(n+2)x^{n}$
Work Step by Step
Since, we know that the Taylor Series for $\dfrac{1}{(1-x)^2}$ is defined as:
$\dfrac{1}{(1-x)^2}=1+2x+3x^2+.....nx^{n-1}$
Now, $ (\dfrac{1}{(1-x)^2})'=[1+2x+3x^2+.....nx^{n-1}]'=2+6x+12x^2+....n(n+1)x^{n-1}$
$=\Sigma_{n=0}^\infty (n+1)(n+2)x^{n}$
