No, you didn't deal with the coefficients right. The sum sign is for addition. The same distributive rules apply to that kind of addition as to any other kind of addition. I would actually move the $\displaystyle 1+x^{2}$ in the $\displaystyle y''$ term into the sum, as well as the $\displaystyle x$ into the $\displaystyle y'$ term thus:

$\displaystyle \displaystyle{\sum_{k=0}^{\infty}(k+2)(k+1)c_{k+2} (x^{k}-x^{k+2})-6\,\sum_{k=0}^{\infty}(k+1)c_{k+1}x^{k+1}-4\,\sum_{k=0}^{\infty}c_{k}x^{k}=0}.$

Now, I think you can add everything together. How does that look?