Answer
$\Sigma_{n=0}^\infty 2^{(n)} x^{(n+2)}$
Work Step by Step
Re-write as: $\dfrac{1}{2-x}=\dfrac{1}{2} [\dfrac{1}{1-\dfrac{x}{2}}]$
Consider the Maclaurin Series for $\dfrac{1}{1-x}$ is defined as:
$\dfrac{1}{1-x}=\Sigma_{n=0}^\infty x^n=1+x+x^2+x^3+x^4+...$
Then, we have $\dfrac{1}{1-2x}=1+(2x)+(2x)^2+(2x)^3+(2x)^4+...$
Now, $\dfrac{x^2}{1-2x}=x^2[1+(2x)+(2x)^2+(2x)^3+(2x)^4+...]$
$\implies \Sigma_{n=0}^\infty 2^{(n)} x^{(n+2)}$
