Let the given statement be $P(n)$ i.e., $P(n)$ :
$\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+\ldots+\frac{1}{(3 n-2)(3 n+1)}=\frac{n}{(3 n+1)}$
For $n=1, \mathrm{LHS}=\frac{1}{1.4}=\frac{1}{4}$ and R.H.S
$=\frac{1}{3.1+1}=\frac{1}{4}$
$\therefore \quad P(n)$ is true for $n=1$
Assume $P(k)$ is true for $n=k$
$\begin{aligned}
&\therefore \frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+\ldots \ldots+\frac{1}{(3 k-2)(3 k+1)} \\
&=\frac{k}{3 k+1} \quad \ldots(i)
\end{aligned}$
Now we have to prove that $P(k+1)$ is true whenever $P(k)$ true.
$P(k+1): \frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+\ldots \ldots$
$+\frac{1}{(3 k-2)(3 k+1)}+\frac{1}{(3 k+1)(3 k+4)}$
$=\frac{k}{3 k+1}+\frac{1}{(3 k+1)(3 k+4)}$
$=\frac{1}{3 k+1}\left[k+\frac{1}{(3 k+4)}\right] \quad$ [from(i) $]$
$=\frac{1}{3 k+1}\left[\frac{k(3 k+4)+1}{3 k+4}\right]=\frac{3 k^2+4 k+1}{(3 k+1)(3 k+4)}$
$=\frac{(k+1)(3 k+1)}{(3 k+1)(3 k+4)}=\frac{k+1}{(3 k+4)}=\frac{\overline{k+1}}{3(\overline{k+1})+1}$
$\therefore \quad P(k+1)$ is true for $n=(k+1)$ i.e., $P(k+1)$ is true whenever $P(k)$ is true.
Hence, by principle of mathematical Induction $P(n)$ is true for all $n \in N$