Answer
$e^{-x^{4}}$
Work Step by Step
Given: $\Sigma_{n=0}^{\infty}{(-1)^{n}}\dfrac{x^{4n}}{n!}$
As we know
$e^{x}=\Sigma_{n=0}^{\infty}\dfrac{x^{n}}{n!}$
and $e^{-x}=\Sigma_{n=0}^{\infty}(-1)^{n}\dfrac{x^{n}}{n!}$
Thus,
$e^{-x^{4}}=\Sigma_{n=0}^{\infty}{(-1)^{n}}\dfrac{x^{4n}}{n!}$
