Question

Prove that for every positive integer n,

$\begin{array}{r}\mathrm{1.2.3}+\mathrm{2.3.4}+\dots +n(n+1)(n+2)\\ =n(n+1)(n+2)(n+3)/4\end{array}$

Short answer

$\begin{array}{r}\mathrm{1.2.3}+\mathrm{2.3.4}+\dots +n(n+1)(n+2)\\ =n(n+1)(n+2)(n+3)/4\end{array}$

Step-by-step solution

Step: 1

Basic step: so, for n=1,

$\begin{array}{r}P\left(1\right)=\mathrm{1.2.3}=\frac{1(1+1)(1+2)(1+3)}{4}\\ 6=6\end{array}$

True for n=1.

Step: 2

Assume for n=k,

$\begin{array}{r}P\left(k\right)=\mathrm{1.2.3}+\mathrm{2.3.4}+\dots +k(k+1)(k+2)\\ =\frac{k(k+1)(k+2)(k+3)}{4}\end{array}$

Step: 3

To prove for n=k+1

$\begin{array}{r}P\left(k\right)=\mathrm{1.2.3}+\mathrm{2.3.4}+\dots +k(k+1)(k+2)+(k+1)(k+2)(k+3)\\ =\frac{k(k+1)(k+2)(k+3)}{3}+(k+1)(k+2)(k+3)\\ =\frac{(k+1)(k+2)(k+3)}{4}(k+4)\\ =\frac{(k+1)(k+2)(k+3)(k+4)}{4}\end{array}$

True for n=k+1.