The given series is $(1\times 2\times 3)+(2\times 3\times 4)+(3\times 4\times 5).......$
Step 1
We have to find the $n^{th}$ term of the series $t_n$.
The $n^{th}$ term is $n^{th}$ bracket.
Each bracket consists of product of $3$ terms.
The first term in each bracket forms a sequence $(1,2,3,......n)$
$\therefore $ The first term in $n^{th}$ bracket $=n$
The $2^{nd}$ term in each bracket forms a sequence $(2,3,4.....n+1)$
$\therefore$ The $2^{nd}$ term in $n^{th}$ bracket $=n+1$
Similarly the $3^{rd}$ term in each bracket are $(3,4,5........n+2)$
$\Rightarrow\:3^{rd}$ term in $n^{th}$ bracket $=n+2$
$\therefore n^{th}$ term of the series $=t_n=n(n+1)(n+2)$
$i.e., t_n=n^3+3n^2+2n$
Step 2
We know that sum of $n$ terms of any series $S_n=\sum t_n$
$\Rightarrow\:S_n=\sum t_n=\sum (n^3+3n^2+2n)$
$=\sum n^3+\sum 3n^2+\sum 2n$
$\bigg($Since $\sum (A+B+C)=\sum A+\sum B+\sum C\bigg)$
Also $\sum k.A=k.\sum A$ if $k$ is a constant.
$\Rightarrow\:S_n=\sum n^3+3.\sum n^2+2.\sum n$
We know that $\sum n^3=\large\frac{n^2(n+1)^2}{4}$, $\sum n^2=\large\frac{n(n+1)(2n+1)}{6}$
and $\sum n=\large\frac{n(n+1)}{2}$
$\Rightarrow\:S_n=\large\frac{n^2(n+1)^2}{4}$$+3.\large\frac{n(n+1)(2n+1)}{6}$$+2.\large\frac{n(n+1)}{2}$
Taking $n(n+1)$ common from each term,
$\Rightarrow\:S_n=n(n+1)\bigg[\large\frac{n(n+1)}{4}+\frac{2n+1}{2}+$$1\bigg]$
$\Rightarrow\:S_n=n(n+1)\bigg[\large\frac{n^2+n+4n+2+4}{4}\bigg]$
$\Rightarrow\:S_n=n(n+1)\bigg[\large\frac{n^2+5n+6}{4}\bigg]$
Factorising $n^2+5n+6=n^2+3n+2n+6$
$=n(n+3)+2(n+3)=(n+2)(n+3)$
$\Rightarrow\:$ Sum of $n$ terms of the series $=S_n=\large\frac{n(n+1)(n+2)(n+3)}{4}$