Let $P(n)$ be the given statement i.e. $P(n): 1.3+3.5+5.7+\ldots \ldots+(2 n-1)(2 n+1)$ $=\frac{n\left(4 n^2+6 n-1\right)}{3}$
Putting $n=1, \mathrm{LHS}=1.3=3$ and
RHS $=\frac{1\left(4.1^2+6.1-1\right)}{3}=\frac{4+6-1}{3}=\frac{9}{3}=3$
$\mathrm{LHS}=\mathrm{RHS}$
$\therefore \quad P(n)$ is true for $n=1$
Assume that $P(n)$ is true for $n=k$
i.e. $P(k): 1.3+3 \cdot 5+5 \cdot 7+\ldots .+(2 k-1)(2 k+1)$
$=\frac{k\left(4 k^2+6 k-1\right)}{3} \quad \ldots(i)$
Now we have to prove that $P(k+1)$ is true whenever $P(k)$ is true.
$\begin{aligned}
&\therefore P(k+1)=1.3+3.5+\ldots \ldots .+(2 k-1)(2 k+1) \\
&+\{2(k+1)-1\}\{2(k+1)+1\} \\
&=\frac{k\left(4 k^2+6 k-1\right)}{3}+(2 k+1)(2 k+3)[\text { from (i) }] \\
&=\frac{\left(4 k^3+6 k^2-k\right)+3(2 k+1)(2 k+3)}{3} \\
&=\frac{4 k^3+18 k^2+23 k+9}{3}=\frac{(k+1)\left(4 k^2+14 k+9\right)}{3} \\
&=\frac{(k+1)\left(4(k+1)^2+6(k+1)-1\right)}{3}
\end{aligned}$
Thus, $P(n)$ is true for $n=k+1$
$\therefore P(k+1)$ is true whenever $P(k)$ is true
Hence, by Principle of mathematical induction $P(n)$ is true for all $n \in N$