Answer
$1-\dfrac{x}{2}+\dfrac{x^2}{8}-\dfrac{x^3}{48}+\dfrac{x^4}{384}-...$
Work Step by Step
Consider the Maclaurin Series for $e^x$ as follows:
$e^x=1+x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\dfrac{x^4}{4!}+...$
Plug $-\dfrac{x}{2}$ instead of $x$ in the above series.
Now, $e^{-\dfrac{x}{2}}=1+(\dfrac{-x}{2})+\dfrac{(-\dfrac{x}{2})^2}{2!}+\dfrac{(-\dfrac{x}{2})^3}{3!}-\dfrac{(-\dfrac{x}{2})^4}{4!}...$
$ \implies 1-\dfrac{x}{2}+\dfrac{x^2}{8}-\dfrac{x^3}{48}+\dfrac{x^4}{384}-...$
