Power Series Solution of $y''+y=0$, and Summation Indices

From your work it follows inductively that for $n\geq0$, $$c_{2n}=\frac{(-1)^n c_0}{(2n)!}, \qquad c_{2n+1}=\frac{(-1)^n c_1}{(2n+1)!} .$$ So we find that $$\begin{align}y(x)&=\sum_{n=0}^\infty c_nx^n =\sum_{n=0}^\infty c_{2n}x^{2n}+\sum_{n=0}^\infty c_{2n+1}x^{2n+1}\\ &=c_0\sum_{n=0}^\infty \frac{(-1)^nx^{2n}}{(2n)!}+c_1\sum_{n=0}^\infty \frac{(-1)^nx^{2n+1}}{(2n+1)!}=c_0\cos(x)+c_1\sin(x).\end{align}$$

Your attempt looks correct to me. You can use the Taylor series definition of: $$\cos (x)=\sum_{n=0}^\infty (-1)^n\dfrac {x^{2n}}{(2n)!}$$ $$\sin (x)=\sum_{n=0}^\infty (-1)^n\dfrac {x^{2n+1}}{(2n+1)!}$$