Answer
$1-\dfrac{x}{2}+\dfrac{1}{8}x^2-\dfrac{1}{48}x^3+\dfrac{1}{384}x^4-...$
Work Step by Step
Since, we know that the Maclaurin Series for $e^x$ is defined as:
$e^x=\Sigma_{n=0}^\infty \dfrac{x^n}{n!}=1+x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\dfrac{x^4}{4!}+...$
Now, $e^{-x/2}=\Sigma_{n=0}^\infty \dfrac{(-x/2)^n}{n!}=1+(\dfrac{-x}{2})+\dfrac{(-x/2)^2}{2!}+\dfrac{(-x/2)^3}{3!}-\dfrac{(-x/2)^4}{4!}...$
or, $=1-\dfrac{x}{2}+\dfrac{1}{8}x^2-\dfrac{1}{48}x^3+\dfrac{1}{384}x^4-...$
