Let the given statement be $P(\mathrm{n})$ i.e.
$P(n): 1^2+3^2+5^2+\ldots+(2 n-1)^2$
$=\frac{n(2 n-1)(2 n+1)}{3}$
For $n=1$, L.H.S. $=1^2=1$
R.H.S. $=\frac{1 .(2-1)(2+1)}{3}=\frac{1.1 \cdot 3}{3}=1$
$P(n)$ is true for $n=1$
Suppose $P(k)$ is true for $n=k$ i.e.
$\begin{aligned}
&1^2+3^2+5^2+\ldots+(2 k-1)^2=\frac{k(2 k-1)(2 k+1)}{3} \\
&\quad k^{\text {th }} \text { term }=(2 k-1)^2 \\
&\therefore(k+1)^{\text {th }} \text { term }=(2(k+1)-1)^2=(2 k+1)^2 \\
&\text { Adding }(2 k+1)^2 \text { to both sides. } \\
&\text { L.H.S. }=1^2+3^2+5^2+\ldots+(2 k-1)^2+(2 k+1)^2 \\
&\text { R.H.S. }=\frac{k(2 k-1)(2 k+1)}{3}+(2 k+1)^2
\end{aligned}$
$\begin{aligned}
&=(2 k+1)\left[\frac{k(2 k-1)}{3}+(2 k+1)\right] \\
&=(2 k+1)\left[\frac{k(2 k-1)+3(2 k+1)}{3}\right] \\
&=(2 k+1)\left(\frac{2 k^2+5 k+3}{3}\right] \\
&=\frac{(2 k+1)(k+1)(2 k+3)}{3} \\
&=\frac{(\overline{k+1})[2(\overline{k+1})-1][2(\overline{k+1})+1]}{3}
\end{aligned}$
Thus $P(k+1)$ is true for $n=k$ i.e.
$P(k+1)$ is true whenever $P(k)$ is true
By principle of mathematical induction $P(n)$ is true for all values of $n \in \mathrm{N}$.