1.2.3 + 2.3.4 + 3.4.5 + ..... n terms =

$\begin{array}{rcl} & & 1.2.3+2.3.4+3.4.5+........\\ {\mathrm{t}}_{\mathrm{n}}& =& \mathrm{n}(\mathrm{n}+1) (\mathrm{n}+2)\\ & =& {\mathrm{n}}^3+3{\mathrm{n}}^2+2\mathrm{n}\\ {\mathrm{S}}_{\mathrm{n}}& =& \sum {\mathrm{t}}_{\mathrm{n}}=\sum \left({\mathrm{n}}^3+3{\mathrm{n}}^2+2\mathrm{n}\right)\\ & =& \frac{{\mathrm{n}}^2{(\mathrm{n}+1)}^2}{4}+\frac{3\mathrm{n}(\mathrm{n}+1) (2\mathrm{n}+1)}{6}+\frac{2\mathrm{n}(\mathrm{n}+1)}{2}\\ & =& \frac{\mathrm{n}(\mathrm{n}+1)}{2} \left[\frac{\mathrm{n}(\mathrm{n}+1)}{2}+2\mathrm{n}+1+2\right]\\ & =& \frac{\mathrm{n}(\mathrm{n}+1)}{2} \left[\frac{{\mathrm{n}}^2+5\mathrm{n}+6}{2}\right]\\ & =& \frac{\mathrm{n}(\mathrm{n}+1) (\mathrm{n}+2) (\mathrm{n}+3)}{4}\end{array}$