Answer
$\dfrac{1}{1-2x}=\Sigma_{n=0}^\infty 2^n x^n $ for $|x| \lt \dfrac{1}{2} $
Work Step by Step
Since, we know that the Taylor Series for $\dfrac{1}{1-x}$ is defined as:
$\dfrac{1}{1-x}=\Sigma_{n=0}^\infty x^n =1+x+x^2+....+x^n$
Replace $x=2x$ in the above form:
$\dfrac{1}{1-2x}=\Sigma_{n=0}^\infty (2x)^n =1+(2x)+(2x)^2+....+(2x)^n$
or, $\dfrac{1}{1-2x}=\Sigma_{n=0}^\infty 2^n x^n $
Here, $|2x| \lt 1 \implies |x| \lt \dfrac{1}{2} $
Hence, $\dfrac{1}{1-2x}=\Sigma_{n=0}^\infty 2^n x^n $ for $|x| \lt \dfrac{1}{2} $
