Answer
$\sum\limits_{k=0}^{n/2} (-1)^k \dfrac{(x \pi)^{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!} + ..\\
$
$f(x)=cos\pi x$
Differentiate w.r.t x
$f'(x)=-\pi sin\pi x$
$f''(x)=-\pi^2 cos\pi x$
$f'''(x)=\pi^3 sin\pi x$
$f''''(x)=\pi^4 cos\pi x$
Now $f(0)=1$, $f'(0)x =0 $, $f''(0)=-\pi^2$, $f'''(0)=0 $, $f''''(x)=\pi^4$
Hence, the Maclaurin Series for the function is
$\displaystyle 1- \frac {\pi^2 x^2} {2!} + \frac {\pi^4 x^4} {4!} - \frac {\pi ^6 x^6 } {6!} + ... (-1)^n\frac{(x \pi)^{2n}} {(2n)!} = \sum\limits_{k=0}^{n/2} (-1)^k \dfrac{(x \pi)^{2k}} {(2k)!} $
