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