>>973
k(k+1)(2k+1)=k(k+1){(k+1)^2-k^2}=k(k+1)^3-k^3(k+1)=k(k+1)^3-(k-1)k^3-2k^3

Σ[k=1,n]k(k+1)(2k+1) = n(n+1)^3-(1/2)n^2(n+1)^2 = (1/2)n(n+1)^2(n+2)