Solution:

Let P(n) be the given statement.

Now,

$P(n)=1^{2}+3^{2}+5^{2}+\ldots+(2 n-1)^{2}=\frac{1}{3} n\left(4 n^{2}-1\right)$

Step 1:

$P(1)=1^{2}=1=\frac{1}{3} \times 1 \times(4-1)$

Hence, $P(1)$ is true.

Step 2 :

Let $P(m)$ be true.

Then,

$1^{2}+3^{2}+\ldots+(2 m-1)^{2}=\frac{1}{3} m\left(4 m^{2}-1\right)$

To prove : $P(m+1)$ is true whenever $P(m)$ is true.

That $i s$

$1^{2}+3^{2}=\ldots+(2 m+1)^{2}=\frac{1}{3}(m+1)\left\{4(m+1)^{2}-1\right\}$

We know that $P(m)$ is true.

Thus, we have :

$1^{2}+3^{2}+\ldots+(2 m-1)^{2}=\frac{1}{3} m\left(4 m^{2}-1\right)$

$\Rightarrow 1^{2}+3^{2}+\ldots$$+(2 m-1)^{2}+(2 m+1)^{2}=\frac{1}{3} m\left(4 m^{2}-1\right)+(2 m+1)^{2}$

$\left[\right.$ Adding $(2 m+1)^{2}$ to both sides $]$

$\Rightarrow P(m+1)=\frac{1}{3}\left(4 m^{3}-m+12 m^{2}+12 m+3\right)$

$\Rightarrow P(m+1)=\frac{1}{3}\left(4 m^{3}-m+8 m^{2}+4 m+4 m^{2}+8 m+3\right)$

$=\frac{1}{3}(m+1)\left(4 m^{2}+8 m+3\right)$

$=\frac{1}{3}(m+1)\left(4(m+1)^{2}-1\right)$

Thus, $P(m+1)$ is true.

$B y$ the $p$ rinciple of $m$ athematical $i$ nduction, $P(n)$ is true for all $\mathrm{n} \in N$.