Answer
$\Sigma_{n=0}^\infty \dfrac{(\pi)^n x^n}{2^{(n)}n!}$
Work Step by Step
Consider the Taylor Series for $e^x$ as follows:
$e^x=\Sigma_{n=0}^\infty \dfrac{x^n}{n!}=1+x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\dfrac{x^4}{4!}+...$
Then $e^{(\dfrac{\pi x}{2})}=\Sigma_{n=0}^\infty \dfrac{(\dfrac{\pi x}{2})^n}{n!}=1+(\dfrac{\pi x}{2})+\dfrac{(\dfrac{\pi x}{2})^2}{2!}+\dfrac{(\dfrac{\pi x}{2})^3}{3!}+\dfrac{(\dfrac{\pi x}{2})^4}{4!}+...$
Hence, $e^{(\dfrac{\pi x}{2})}=\Sigma_{n=0}^\infty \dfrac{(\pi)^n x^n}{2^{(n)}n!}$
