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