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

Let the given statement be $P(n)$, i.e.,
$P(n) : 1.3+3.5+5.7+...+(2n-1)(2n+1)= \large\frac{n(4n^2+6n-1)}{3}$
For $n = 1$, we have
$P(1):1.3=3= \large\frac{1(4.1^2+6.1-1)}{3} $$= \large\frac{4+6-1}{3}$$=\large\frac{9}{3}$$= 3, which is true. Let P(k) be true for some positive integer k, i.e., = 1.3+3.5+5.7+...+(2k-1)(2k+1)= \large\frac{k(4k^2+6k-1)}{3}-----------(i) We shall now prove that P(k+1) is true. Consider (1.3+3.5+5.7+...+(2k-1)(2k+1)+ \{2(k+1)-1 \} \{ 2(k+1)+1 \} = \large\frac{k(4k^2+6k-1)}{3}$$+(2k+2-1)(2k+2+1)\qquad$ [ Using (i) ]
$= \large\frac{k(4k^2+6k-1)}{3} $$+(2k+1)(2k+3) = \large\frac{k(4k^2+6k-1)}{3}$$(4k^2+8k+3)$
$= \large\frac{k(4k^2+6k-1)+3(4k^2+8k+3)}{3}$
$= \large\frac{4k^3+6k^2-k+12k^2+24k+9}{3}$
$= \large\frac{4k^3+18k^2+23k+9}{3}$
$= \large\frac{4k^3+14k^2+9k+4k^2+14k+9}{3}$
$= \large\frac{k(4k^2+14k+9)+1(4k^2+14k+9)}{3}$
$= \large\frac{(k+1)4k^2+14k+9}{3}$
$= \large\frac{(k+1) \{4k^2+8k+4+6k+6-1\}}{3}$
$= \large\frac{(k+1) \{ 4(k^2+2k+1) +6(k+1)-1\}}{3}$
$= \large\frac{(k+1) \{ 4(k+1)^2+6(k+1)-1\}}{3}$
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 i.e., $n$.