Answer
$\dfrac{6x^2}{(3-x)^3}$
Work Step by Step
The power series of $(1-x)^{-3}$ can be written as:
$1+3x+4x^2+10x^3+......=\Sigma_{k=0}^{\infty} \dbinom{k+2}{k}x^k$
Multiply the above equation by $2$ and replace $x$ by $\dfrac{x}{3}$ in the above series to obtain:
$\dfrac{2x^2}{3^2}(1+3(\dfrac{(x}{3})+6(\dfrac{x}{3})^3+......=\dfrac{2x^2}{3^2}(1-\dfrac{x}{3})^{-3}$
Therefore, the function represented by the power series is:
$\Sigma_{k=1}^{\infty} \dfrac{k(k-1)x^k}{3^k}=\dfrac{6x^2}{(3-x)^3}$
