# Prove the following by using the principle of mathematical induction for all $n\in \mathbb{N}$ :Q : 7        $1.3+3.5+5.7+...+(2n-1)(2n+1)=\frac{n(4n^2+6n-1)}{3}$

Let the given statement be p(n) i.e.
$p(n):1.3+3.5+5.7+...+(2n-1)(2n+1)=\frac{n(4n^2+6n-1)}{3}$
For n = 1  we have
$p(1):1.3=3=\frac{1(4(1)^2+6(1)-1)}{3}= \frac{4+6-1}{3}=\frac{9}{3}=3$   ,   which is true

For  n = k  we have

$p(k):1.3+3.5+5.7+...+(2k-1)(2k+1)=\frac{k(4k^2+6k-1)}{3} \ \ \ \ \ \ \ \ -(i)$   ,        Let's assume that this statement is true

Now,
For  n = k + 1  we have
$p(k+1):1.3+3.5+5.7+...+(2(k+1)-1)(2(k+1)+1)$                                                                                                                                                                                                                      $=1.3+3.5+5.7+...+(2k-1)(2k+1)+(2(k+1)-1)(2(k+1)+1)$

$=\frac{k(4k^2+6k-1)}{3}+(2k+1)(2k+3) \ \ \ \ \ \ \ \ \ \ (using \ (i))$
$=\frac{k(4k^2+6k-1)+3(2k+1)(2k+3)}{3}$
$=\frac{k(4k^2+6k-1)+3(4k^2+8k+3)}{3}$
$=\frac{(4k^3+6k^2-k+12k^2+28k+9)}{3}$
$=\frac{(4k^3+18k^2+23k+9)}{3}$
$=\frac{(4k^3+14k^2+9k+4k^2+14k+9)}{3}$
$=\frac{(k(4k^2+14k+9)+4k^2+14k+9)}{3}$
$=\frac{(4k^2+14k+9)(k+1)}{3}$
$=\frac{(k+1)(4k^2+8k+4+6k+6-1)}{3}$
$=\frac{(k+1)(4(k^2+2k+1)+6(k+1)-1)}{3}$
$=\frac{(k+1)(4(k+1)^2+6(k+1)-1)}{3}$

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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