# Prove the following by using the principle of mathematical induction for all $n\in \mathbb{N}$ :Q : 15        $1^2+3^2+5^2+...+(2n-1)^2=\frac{n(2n-1)(2n+1)}{3}$

Let the given statement be p(n) i.e.
$p(n):1^2+3^2+5^2+...+(2n-1)^2=\frac{n(2n-1)(2n+1)}{3}$
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

For  n = k  we have

$p(k):1^2+3^2+5^2+...+(2k-1)^2=\frac{k(2k-1)(2k+1)}{3} \ \ \ \ \ \ \ \ \ \ -(i)$   ,        Let's assume that this statement is true

Now,
For  n = k + 1  we have
$p(k+1):1^2+3^2+5^2+...+(2(k+1)-1)^2$                                                                                                                                                                                                                                                    $=1^2+3^2+5^2+...+(2k-1)^2+(2(k+1)-1)^2$

$=\frac{k(2k-1)(2k+1)}{3}+(2(k+1)-1)^2 \ \ \ \ \ \ \ \ \ \ (using \ (i))$
$=\frac{k(2k-1)(2k+1)+3(2(k+1)-1)^2}{3}$
$=\frac{k(2k-1)(2k+1)+3(2k+1)^2}{3}$
$=\frac{(2k+1)(k(2k-1)+3(2k+1))}{3}$
$=\frac{(2k+1)(2k^2-k+6k+3)}{3}$
$=\frac{(2k+1)(2k^2+5k+3)}{3}$
$=\frac{(2k+1)(2k^2+2k+3k+3)}{3}$
$=\frac{(2k+1)(2k+3)(k+1)}{3}$
$=\frac{(k+1)(2(k+1)-1)(2(k+1)+1)}{3}$

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 n

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