Let $P(n)$ be the given statement i.e.
$\begin{aligned}
&P(n): \frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\ldots+\frac{1}{(2 n+1)(2 n+3)} \\
&=\frac{n}{3(2 n+3)}
\end{aligned}$
For $n=1$, L.H.S. $=\frac{1}{3.5}=\frac{1}{15}$
$\text { R.H.S. }=\frac{1}{3 \cdot(2+3)}=\frac{1}{3 \cdot 5}=\frac{1}{15}$
Suppose $P(k)$ be true for $n=k$ i.e.
$\begin{aligned}
&\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\ldots+\frac{1}{(2 k+1)(2 k+3)}=\frac{k}{3(2 k+3)} \\
&k^{\text {th }} \text { term }=\frac{1}{(2 k+1)(2 k+3)} \\
&(k+1)^{\text {th }} \text { term }=\frac{1}{[2(k+1)+1][2(k+1)+3]}
\end{aligned}$
$=\frac{1}{(2 k+3)(2 k+5)}$
Adding $\frac{1}{(2 k+3)(2 k+5)}$ to both sides,
$\text { L.H.S. }=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\ldots+\frac{1}{(2 k+1)(2 k+3)}$
$+\frac{1}{(2 k+3)(2 k+5)}$
$\begin{aligned}
&\text { R.H.S. }=\frac{k}{3(2 k+3)}+\frac{1}{(2 k+3)(2 k+5)} \\
&=\frac{1}{2 k+3}\left[\frac{k}{3}+\frac{1}{2 k+5}\right] \\
&=\frac{1}{2 k+3}\left[\frac{k(2 k+5)+3}{3(2 k+5)}\right] \\
&=\frac{2 k^2+5 k+3}{3(2 k+3)(2 k+5)}=\frac{(k+1)(2 k+3)}{3(2 k+3)(2 k+5)} \\
&=\frac{\overline{k+1}}{3(2 k+5)}=\frac{k+1}{3[2(\overline{k+1})+3]}
\end{aligned}$
Hence $P(k+1)$ 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 $n \in \mathrm{N}$.