Prove the following by using the principle of mathematical induction for all n ∈ N:

Let the given statement be P(n), i.e.,

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

For $n=1$, we have

$P(1)=1^{2}=1=\frac{1(2.1-1)(2.1+1)}{3}=\frac{1.1 .3}{3}=1$, which is true.

Let $\mathrm{P}(k)$ be true for some positive integer $k$, i.e.,

$\mathrm{P}(k)=1^{2}+3^{2}+5^{2}+\ldots+(2 k-1)^{2}=\frac{k(2 k-1)(2 k+1)}{3}$     (1)

We shall now prove that P(k + 1) is true.

Consider

$\left\{1^{2}+3^{2}+5^{2}+\ldots+(2 k-1)^{2}\right\}+\{2(k+1)-1\}^{2}$

$=\frac{k(2 k-1)(2 k+1)}{3}+(2 k+2-1)^{2}$

$=\frac{k(2 k-1)(2 k+1)}{3}+(2 k+1)^{2}$

$=\frac{k(2 k-1)(2 k+1)+3(2 k+1)^{2}}{3}$

$=\frac{(2 k+1)\{k(2 k-1)+3(2 k+1)\}}{3}$

$=\frac{(2 k+1)\left\{2 k^{2}-k+6 k+3\right\}}{3}$

$=\frac{(2 k+1)\left\{2 k^{2}+5 k+3\right\}}{3}$

$=\frac{(2 k+1)\left\{2 k^{2}+2 k+3 k+3\right\}}{3}$

$=\frac{(2 k+1)(k+1)(2 k+3)}{3}$

$=\frac{(k+1)\{2(k+1)-1\}\{2(k+1)+1\}}{3}$

Thus, P(k + 1) is true whenever P(k) is true.

Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.