Solution:

Let P(n) be the given statement.

Now,

$P(n)=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+\ldots+\frac{1}{(3 n-1)(3 n+2)}=\frac{n}{6 n+4}$

Step 1:

$P(1)=\frac{1}{2.5}=\frac{1}{10}=\frac{1}{6+4}$

Hence, $P(1)$ is true.

Step 2;

Let $P(m)$ be true.

Then,

$\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+\ldots+\frac{1}{(3 m-1)(3 m+2)}=\frac{m}{6 m+4}$

To prove : $P(m+1)$ is true.

i. e.,

$\frac{1}{2.5}+\frac{1}{5.8}+\ldots+\frac{1}{(3 m+2)(3 m+5)}=\frac{m+1}{6 m+10}$

Thus, we have;

$\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+\ldots+\frac{1}{(3 m-1)(3 m+2)}=\frac{m}{6 m+4}$

$\Rightarrow \frac{1}{2.5}+\frac{1}{5.8}+\ldots+\frac{1}{(3 m-1)(3 m+2)}+\frac{1}{(3 m+2)(3 m+5)}=\frac{m}{6 m+4}+\frac{1}{(3 m+2)(3 m+5)}$

$\left[\right.$ Adding $\frac{1}{(3 m+2)(3 m+5)}$ to both sides $]$

$\Rightarrow \frac{1}{2.5}+\frac{1}{5.8}+\ldots+\frac{1}{(3 m+2)(3 m+5)}=\frac{3 m^{2}+5 m+2}{2(3 m+2)(3 m+5)}=\frac{(3 m+2)(m+1)}{2(3 m+2)(3 m+5)}=\frac{m+1}{6 m+10}$

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

By the $p$ rinciple of $m$ athematical induction, $P(n)$ is true for all $n \in N$.