Let P(n) be the given statement.

Now,

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

Step 1:

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

Hence, $P(1)$ is true.

Step 2:

Let $P(m)$ be true.

Then,

$\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\ldots+\frac{1}{m(m+1)}=\frac{m}{m+1}$

We shall now prove that $P(m+1)$ is true.

i. e.,

$\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\ldots+\frac{1}{(m+1)(m+2)}=\frac{m+1}{m+2}$

Now,

$P(m)=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\ldots+\frac{1}{m(m+1)}=\frac{m}{m+1}$

$\Rightarrow \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\ldots+\frac{1}{m(m+1)}+\frac{1}{(m+1)(m+2)}=\frac{m}{m+1}+\frac{1}{(m+1)(m+2)}$

$\left[\right.$ Adding $\frac{1}{(m+1)(m+2)}$ to both sides $]$

$\Rightarrow \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\ldots+\frac{1}{(m+1)(m+2)}=\frac{m^{2}+2 m+1}{(m+1)(m+2)}=\frac{(m+1)^{2}}{(m+1)(m+2)}=\frac{m+1}{m+2}$

Therefore, $P(m+1)$ is true.

By the principle of $m$ athematical $i$ nduction, $P(n)$ is true for all $n \in \mathrm{N}$.