Sum the series $1\cdot2\cdot3 + 2\cdot3\cdot4 + 3\cdot4\cdot5 + \ldots$ to $n$ terms.

The nth term is $$t_n=n(n+1)(n+2)=n^3+3n^2+2n.$$Hence $$S_n=\sum_{k=1}^n (k^3+3k^2+2k)=\sum_{k=1}^n k^3+3\sum_{k=1}^n k^2+2\sum_{k=1}^n k.$$Using the standard sums, $$S_n=\frac{n^2(n+1)^2}{4}+3\cdot \frac{n(n+1)(2n+1)}{6}+2\cdot \frac{n(n+1)}{2}.$$On simplification this becomes $$S_n=\frac{1}{4}n(n+1)(n+2)(n+3).$$