Question

Prove that: $\mathrm{1.2.3}+\mathrm{2.3.4}+\cdots +n(n+1)(n+2)=\frac{n(n+1)(n+2)(n+3)}{12}$

Answer 1

$S=1\times 2\times 3+2\times 3\times 4+3\times 4\times \mathrm{5.......}n(n+1)(n+2)S=\sum _{r=1}^{r=n}r(r+1)(r+1)S=\sum _{r=}^n{r}^3+r^2+rS=\sum _{r=}^n{r}^3+\sum _{r=}^n{r}^2+\sum _{r=}^{n}rS={\left(\frac{n(n+1)}{2}\right)}^2+\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}S=\frac{n^4+n^2+2n^3}{4}+\frac{2n^3+3n^2+n}{6}+\frac{n^2+n}{2}S=\frac{{3(n}^4+n^2+2n^3)+2(2n^3+3n^2+n)+6(n^2+n)}{12}S=\frac{3n^4+10n^3+14n^2+8n}{12}$

$S=\frac{n(n+1)(n+2)(n+3)}{12}$