Let the given statement be $P(n)$ i.e.
$P(n):$
$\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+\ldots+\frac{1}{(3 n-1)(3 n+2)}=\frac{n}{6 n+4}$
Putting $n=1, \mathrm{LHS}=\frac{1}{2.5}=\frac{1}{10}$ and
$\mathrm{RHS}=\frac{1}{6+4}=\frac{1}{10}$
$\therefore \quad P(n)$ is true for $n=1$
Let $P(n)$ is true for $n=k$
i.e $P(k): \frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+\ldots \ldots \ldots$
$+\frac{1}{(3 k-1)(3 k+2)}=\frac{k}{6 k+4} \quad \ldots(i)$
We have to prove that $P(k+1)$ is true whenever $P(k)$ is true.
$\begin{aligned} \therefore P(k+1)=\frac{1}{2.5}+& \frac{1}{5.8}+\frac{1}{8.11}+\ldots \ldots \ldots . \\ &+\frac{1}{(3 k-1)(3 k+2)}+\frac{1}{(3 k+2)(3 k+5)} \end{aligned}$
$=\frac{k}{6 k+4}+\frac{1}{(3 k+2)(3 k+5)}$
$=\frac{k}{2(3 k+2)}+\frac{1}{(3 k+2)(3 k+5)} \quad$ [using (i)]
$=\frac{k(3 k+5)+2}{2(3 k+2)(3 k+5)}=\frac{3 k^2+5 k+2}{2(3 k+2)(3 k+5)}$
$=\frac{(3 k+2)(k+1)}{2(3 k+2)(3 k+5)}=\frac{(k+1)}{2(3 k+5)}$
$=\frac{(k+1)}{(6 k+10)}=\frac{(k+1)}{6(k+1)+4}$
This shows that $P(n)$ is true for $n=k+1$ $\therefore P(k+1)$ is true whenever $P(k)$ is true i.e., By principle of mathematical induction $P(n)$ is true for all $n \in N$.