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