https://www.esaral.com/q/find-the-sum-to-n-terms-of-the-series-1-2-2-3-3-4-4-5-32672/

Solution:

The given series is $1 \times 2 \times 3+2 \times 3 \times 4+3 \times 4 \times 5+\ldots$

$n^{\text {th }}$ term, $a_{n}=n(n+1)(n+2)$

$=\left(n^{2}+n\right)(n+2)$

$=n^{3}+3 n^{2}+2 n$

$\therefore S_{n}=\sum_{k=1}^{n} a_{k}$

$=\sum_{k=1}^{n} k^{3}+3 \sum_{k=1}^{n} k^{2}+2 \sum_{k=1}^{n} k$

$=\left[\frac{n(n+1)}{2}\right]^{2}+\frac{3 n(n+1)(2 n+1)}{6}+\frac{2 n(n+1)}{2}$

$=\left[\frac{n(n+1)}{2}\right]^{2}+\frac{n(n+1)(2 n+1)}{2}+n(n+1)$

$=\frac{n(n+1)}{2}\left[\frac{n(n+1)}{2}+2 n+1+2\right]$

$=\frac{n(n+1)}{2}\left[\frac{n^{2}+n+4 n+6}{2}\right]$

$=\frac{n(n+1)}{4}\left(n^{2}+5 n+6\right)$

$=\frac{n(n+1)}{4}\left(n^{2}+2 n+3 n+6\right)$

$=\frac{n(n+1)[n(n+2)+3(n+2)]}{4}$

$=\frac{n(n+1)(n+2)(n+3)}{4}$