Question

$\mathrm{1.2.3}+\mathrm{2.3.4}+\cdots +n(n+1)(n+2)=\frac{n(n+1)(n+2)(n+3)}{4}$

Answer 1

Let $P\left(n\right)=\mathrm{1.2.3}+\mathrm{2.3.4}+\cdots +n(n+1)(n+2)=\frac{n(n+1)(n+2)(n+3)}{4}$
For n = 1
$P\left(1\right)=1\times 2\times 3=\frac{1\times 2\times 3\times 4}{4}\Rightarrow 6=6\therefore P\left(1\right)$ is true
Let P(n) be true for n = k.
$\therefore P\left(k\right)=\mathrm{1.2.3}+\mathrm{2.3.4}+\cdots +k(k+1)(k+2)=\frac{k(k+1)(k+2)(k+3)}{4}\text{For}n=k+1P(k+1)=\mathrm{1.2.3}+\mathrm{2.3.4}+\cdots +k(k+1)(k+2)+(k+1)(k+2)(k+3)=\frac{k(k+1)(k+2)(k+3)}{4}+(k+1)(k+2)(k+3)=(k+1)(k+2)(k+3)[\frac{k}{4}+1]=(k+1)(k+2)(k+3)\left[\frac{k+4}{4}\right]=\frac{(k+1)(k+2)(k+3)(k+4)}{4}$
$\therefore P(k+1)$ is true.
Thus P (k) is true $\Rightarrow P(k+1)$ is true
Hence by principle of mathematical induction,
P(n) is true for all $nϵN.$