Answer
$\dfrac{1}{2}+\dfrac{x}{4}+\dfrac{x^2}{8}+\dfrac{x^3}{16}+\dfrac{x^4}{32}+...$
Work Step by Step
Since, we know $\dfrac{1}{1-x}=\Sigma_{n=0}^\infty x^n=1+x+x^2+x^3+x^4+...$
Now, the given equation can be written as: $\dfrac{1}{2-x}=\dfrac{1}{2} [\dfrac{1}{1-\dfrac{x}{2}}]$
or, $\dfrac{1}{1-\dfrac{x}{2}}=1+(\dfrac{x}{2})+(\dfrac{x}{2})^2+(\dfrac{x}{2})^3+(\dfrac{x}{2})^4+...$
or, $=\dfrac{1}{2}[1+\dfrac{x}{2}+\dfrac{x^2}{4}+\dfrac{x^3}{8}+\dfrac{x^4}{16}+...]$
or, $=\dfrac{1}{2}+\dfrac{x}{4}+\dfrac{x^2}{8}+\dfrac{x^3}{16}+\dfrac{x^4}{32}+...$
