By expanding (2r+1)^5-(2r-1)^5, derive a formula for $\displaystyle \sum{r^{4}}$ from 1 to n.

I've simplified the equation to 2(80r^4+40r^2+1), where (2r+1)^5-(2r-1)^5 is f(n+1)-f(n-1). I have no slightest inclination on where to continue. anyone care to pint me in the right direction?