Answer
$$ \Sigma_{k=0}^{\infty} \frac {x^{2k}}{(2k)!} $$
Work Step by Step
The Maclaurin Series is
$ \displaystyle\sum\limits_{n=0}^{\infty} \frac{f^n(0)x^n} {n!} = f(0) + f'(0)x + \frac{(f''(0)x^2)}{2!} + ..\\
$
$f(x)=\cos h x=\dfrac{e^x+e^{-x}}{2}$
Differentiate
$f'(x)=\dfrac{e^x-e^{-x}}{2}$
$f''(x)=\dfrac{e^x+e^{-x}}{2}$
$f'''(x)=\dfrac{e^x-e^{-x}}{2}$
$f''''(x)=\dfrac{e^x+e^{-x}}{2}$
Now $f(0)=1$ , $f'(0)=0$ , $f''(0)=1$, $f'''(0)=0$, $f''''(0)=1$
Hence the Maclaurin Series for the function is
$\displaystyle 1+ \frac {x^2} {2} + .......+ \frac {x^{2n}}{(2n)!} = \Sigma_{k=0}^{\infty} \frac {x^{2k}}{(2k)!} $
