Question

$1.2+2.3+3.4+....+n(n+1)=\frac{n(n+1)(n+2)}{3}$

Answer 1

Let P(n) : $1.2+2.3+3.4+...+n(n+1)=\frac{n(n+1)(n+2)}{3}$

For n = 1

$1.2=\frac{1(1+1)(1+2)}{3}$

$2=2$

$\Rightarrow$ P(n) is true for n = 1

Let P(n) is true for n = k

$\Rightarrow 1.2+2.3+3.4+...+k(k+1)=\frac{k(k+1)(k+2)}{3}$ .....(1)

We have to show all,

$1.2+2.3+3.4+....+k(k+1)+(k+1)(k+2)=\frac{(k+1)(k+2)(k+3)}{3}$

Now,

$\{1.2+2.3+3.4+....+k(k+1\left)\right\}+(k+1)(k+2)$

$=\frac{k(k+1)(k+2)}{3}+\frac{(k+1)(k+2)}{1}$

$=(k+1)(k+2)[\frac{k}{3}+1]$

$=\frac{(k+1)(k+2)(k+3)}{3}$

$\Rightarrow$ P(n) is true for n = k + 1

$\Rightarrow$ P(n) is true for all n $epsilon$ N by PMI