$\frac{1^2.2}{1!}+\frac{2^2.3}{2!}+\frac{3^2.4}{3!}+\dots \dots \infty =\dots \dots$

$\begin{array}{rcl} & & \sum _1^{\infty } \frac{{\mathrm{n}}^2(\mathrm{n}+1)}{\overline{)\underline{\mathrm{n}}}}\\ & =& \sum _1^{\infty } \frac{\mathrm{n}(\mathrm{n}+1)}{\overline{)\underline{\mathrm{n}-1}}}\\ & =& \sum _1^{\infty } \frac{{\mathrm{n}}^2+\mathrm{n}}{\overline{)\underline{\mathrm{n}-1}}}\\ & =& \sum _1^{\infty } \frac{({\mathrm{n}}^2-1)+(\mathrm{n}-1)+2}{\overline{)\underline{\mathrm{n}-1}}}\\ & =& \sum _1^{\infty } \left(\frac{\mathrm{n}+1}{\overline{)\underline{\mathrm{n}-2}}}+\frac{1}{\overline{)\underline{\mathrm{n}-2}}}+\frac{2}{\overline{)\underline{\mathrm{n}-1}}}\right)\\ & =& \sum _1^{\infty } \frac{(\mathrm{n}-2)+3}{\overline{)\underline{\mathrm{n}-2}}}+\mathrm{e}+2\mathrm{e}\\ & =& \mathrm{e}+3\mathrm{e}+3\mathrm{e}=7\mathrm{e}\end{array}$