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