Answer
$\Sigma_{n=0}^\infty \dfrac{x^{(2n)}}{(2n)!}$
Work Step by Step
Consider the Maclaurin Series for $\cos x$ and $e^x$ are as follows:
$ \cos x=\Sigma_{n=0}^\infty \dfrac{(-1)^n x^{2n}}{(2n)!}$
and
$e^x=\Sigma_{n=0}^\infty \dfrac{x^n}{n!}=(1+x+\dfrac{x^2}{2!}+...$
Then, $ \cosh x=\dfrac{e^x+e^{-x}}{2}$
$ \implies \dfrac{1}{2}[(1+x+\dfrac{x^2}{2!}+...)+(1-x+\dfrac{x^2}{2!}+...)]=1+\dfrac{1}{2!}x^2+\dfrac{1}{4!} x^4=\Sigma_{n=0}^\infty \dfrac{x^{(2n)}}{(2n)!}$
