Question

Prove 1.2 + 2.3 + 3.4 + $\cdots +n(n+1)=\left[\frac{n(n+1)(n+2)}{3}\right].$

Answer 1

Let $P\left(n\right)=1.2+2.3+3.4+\cdots +n(n+1)=\left[\frac{n(n+1)(n+2)}{3}\right].$
For n = 1
$P\left(1\right)=1(1+1)=\frac{1(1+1)(1+2)}{3}\Rightarrow 2=2$
$\therefore$ P (1) is true
Let P (n) be true for n = k
$\therefore P\left(k\right)=1.2+2.3+3.4+\cdots ++k(k+1)=\left[\frac{k(k+1)(k+2)}{3}\right]\text{For}=k+1P(k+1)=1.2+2.3+3.4+\cdots +k(k+1)+(k+1)(k+2)=\frac{k(k+1)(k+2)}{3}+(k+1)(k+2)=(k+1)(k+2)[\frac{k}{3}+1]=(k+1)(k+2)\left[\frac{k+3}{3}\right]=\frac{(k+1)(k+2)(k+3)}{3}\therefore P(k+1)$ is true
thus P(k) is true $\Rightarrow$ P(k +1) is true
Hence by principle fo mathematical induction,