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