1 × 2 + 2 × 3 + 3 × 4 + 4 × 5 + ...
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:
$S_{n}=\sum_{k=1}^{n} T_{k}$
$\Rightarrow S_{n}=\sum_{k=1}^{n}\left(k^{2}+k\right)$
$\Rightarrow S_{n}=\sum_{k=1}^{n} k^{2}+\sum_{k=1}^{n} k$
$\Rightarrow S_{n}=\left(\frac{n(n+1)(2 n+1)}{6}+\frac{n(n+1)}{2}\right)$
$\Rightarrow S_{n}=\frac{n(n+1)}{2}\left(\frac{2 n+1}{3}+1\right)$
$\Rightarrow S_{n}=\frac{n(n+1)}{2}\left(\frac{2 n+4}{3}\right)$
$\Rightarrow S_{n}=\frac{n(n+1)(2 n+4)}{6}$
$\Rightarrow S_{n}=\frac{n(n+1)(n+2)}{3}$
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