1 × 2 + 2 × 3 + 3 × 4 + 4 × 5 + ...

Solution:

Let $T_{n}$ be the $n$th term of the given series.

Thus, we have:

$T_{n}=n(n+1)=n^{2}+n$

Now, let $S_{n}$ be the sum of $n$ terms of the given series.

Thus, we have:

$S_{n}=\sum_{k=1}^{n} T_{k}$

$\Rightarrow S_{n}=\sum_{k=1}^{n}\left(k^{2}+k\right)$

$\Rightarrow S_{n}=\sum_{k=1}^{n} k^{2}+\sum_{k=1}^{n} k$

$\Rightarrow S_{n}=\left(\frac{n(n+1)(2 n+1)}{6}+\frac{n(n+1)}{2}\right)$

$\Rightarrow S_{n}=\frac{n(n+1)}{2}\left(\frac{2 n+1}{3}+1\right)$

$\Rightarrow S_{n}=\frac{n(n+1)}{2}\left(\frac{2 n+4}{3}\right)$

$\Rightarrow S_{n}=\frac{n(n+1)(2 n+4)}{6}$

$\Rightarrow S_{n}=\frac{n(n+1)(n+2)}{3}$