The values of the sum 1.2.3+2.3.4+3.4.5+……. Up to n terms

Put $n=1$

Options verification.

or 

$\text{ General term is }{T}_r=r(r+1)(r+2)=r\left(r^2+3r+2\right)=r^3+3r^2+2r$

$\text{ Hence required sum is }=\sum _{\mathrm{r}=1}^{\mathrm{n}} \left({\mathrm{r}}^3+3{\mathrm{r}}^2+2\mathrm{r}\right)$

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

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

$\begin{array}{l}=\frac{1}{4}n(n+1)\left[n\right(n+1)+2(2n+1)+4]\end{array}$

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

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