Answer
$f(x)+f''(x)=0$
Work Step by Step
$f(x)=\Sigma_{n=0}^{\infty}(-1)^{n}\frac{x^{2n}}{2n}$,
$f'(x)=\Sigma_{n=1}^{\infty}(-1)^{n}\frac{2nx^{2n-1}}{(2n)!}$
$f''(x)=\Sigma_{n=1}^{\infty}(-1)^{n}\frac{2n(2n-1)x^{2n-2}}{(2n)!}$
Let $f(x)+f''(x)=I$
Then $I=\Sigma_{n=0}^{\infty}(-1)^{n}\frac{x^{2n}}{2n}+\Sigma_{n=1}^{\infty}(-1)^{n}\frac{2n(2n-1)x^{2n-2}}{(2n)!}$
After simplification
$f''(x)=-f(x)$
Thus, $f(x)+(-f(x))=0$
or
Hence, $f(x)+f''(x)=0$
