Answer
$\dfrac{1}{2}+\dfrac{x}{4}+\dfrac{1}{8}x^2+\dfrac{1}{16}x^3+\dfrac{1}{32}x^4+...$
Work Step by Step
Re-write the equation as: $\dfrac{1}{2-x}=\dfrac{1}{2} [\dfrac{1}{1-\dfrac{x}{2}}]$
Consider the Taylor's Series for $\dfrac{1}{1-x}$ as follows:
$\dfrac{1}{1-x}=1+x+x^2+x^3+x^4+...$
Plug $(\dfrac{x}{2})$ instead of $x$ in the above series as follows:
$\dfrac{1}{1-\dfrac{x}{2}}=1+(\dfrac{x}{2})+(\dfrac{x}{2})^2+(\dfrac{x}{2})^3+(\dfrac{x}{2})^4+...$
This implies that $\dfrac{1}{2}[1+\dfrac{x}{2}+\dfrac{x^2}{4}+\dfrac{x^3}{8}+\dfrac{x^4}{16}+...]=\dfrac{1}{2}+\dfrac{x}{4}+\dfrac{1}{8}x^2+\dfrac{1}{16}x^3+\dfrac{1}{32}x^4+...$
