#### Answer

$\Sigma_{n=0}^{\infty}(-1)^{n}\frac{x^{4n+1}}{2^{2n}(2n)!}$

#### Work Step by Step

$xcos(\frac{1}{2}x^{2})=x\Sigma_{n=0}^{\infty}(-1)^{n}\frac{(\frac{1}{2}x^{2})^{2n}}{2n!}$
$=x\Sigma_{n=0}^{\infty}(-1)^{n}\frac{(\frac{1}{2})^{2n}x^{4n}}{2n!}$
$=\Sigma_{n=0}^{\infty}(-1)^{n}\frac{x^{4n+1}}{2^{2n}(2n)!}$