If the nth term of a series is written as $n(n + 1)$, find the sum of the first $n$ terms.

We have $t_n=n(n+1)=n^2+n$. Hence $$S_n=\sum_{k=1}^n (k^2+k)=\sum_{k=1}^n k^2+\sum_{k=1}^n k.$$Using $\sum_{k=1}^n k^2=\frac{n(n+1)(2n+1)}{6}$ and $\sum_{k=1}^n k=\frac{n(n+1)}{2}$, $$S_n=\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}=\frac{n(n+1)(n+2)}{3}.$$