Question

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

Answer 1

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

For n = 1

$p\left(1\right):\frac{1}{1.2}=\frac{1}{1+1}$

$\frac{1}{2}=\frac{1}{2}$

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

Let P(n) is true for n = k, so

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

We have to show that

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

Now,

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

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

[Using equation (1)]

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

$=\frac{1}{k+1}\left[\frac{k^2+2k+1}{(k+2)}\right]$

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

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

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

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