Answer
$\Sigma_{n=0}^\infty 2^{(n)} x^{(n)} $ for $|x| \lt \dfrac{1}{2}$
Work Step by Step
Consider the Taylor Series for $\dfrac{1}{1-x}$ as follows:
$\dfrac{1}{1-x}=\Sigma_{n=0}^\infty x^n =1+x+x^2+....+x^n$
and $\dfrac{1}{1-2x}=1+(2x)+(2x)^2+....+(2x)^n=\Sigma_{n=0}^\infty 2^n x^n $
As we can see that $|2x| \lt 1 \implies |x| \lt \dfrac{1}{2} $
Thus, we have $\dfrac{1}{1-2x}=\Sigma_{n=0}^\infty 2^{(n)} x^{(n)} $ for $|x| \lt \dfrac{1}{2} $
