Simplifying sum involving combinations

Question

One of the steps in my textbook is $$\sum_{k=1}^nk(k-1)+k\binom{2n-2k}{2}=2\sum_{k=1}^nk^3-(4n-2)\sum_{k=1}^nk^2+(2n^2-n-1)\sum_{k=1}^nk$$

I do not know how to go from LHS to the RHS

Answer 1

Expand the binomial $$\begin {align} k(k-1)+k\binom{2n-2k}{2}&=k(k-1)+\frac 12k(2n-2k)(2n-2k-1)\\ &=k^2-k+2n^2k-4nk^2+2k^3-nk-k^2\\ &=2k^3-(4n-2)k^2+(2n^2-n-1)k \end {align}$$