Question

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

Answer 1

Let the given statement be $P\left(n\right)$, i.e.,
$P\left(n\right):1.3+3.5+5.7+.......+(2n-1)(2n+1)=\frac{n(4n^2+6n-1)}{3}$
For $n=1$, we have
$P\left(1\right):1.3=3=\frac{1(4.1^2+6.1-1)}{3}=\frac{4+6+1}{3}=\frac{9}{3}=3$, which is true
Let $P\left(k\right)$ be true for some $k\in N$, i.e.,
$1.3+3.5+5.7+......+(2k-1)(2k+1)=\frac{k(4k^2+6k-1)}{3}.........\left(i\right)$
We shall now prove that $P(k+1)$ is true. i.e.$1.3+3.5+5.7+.......+(2k-1)(2k+1)+\left\{2\right(k+1)-1\}\left\{2\right(k+1)+1\}=\frac{(k+1)\{4{(k+1)}^2+6(k+1)-1\}}{3}$
Consider LHS,
$(1.3+3.5+5.7+.......+(2k-1\left)\right(2k+1)+\{2(k+1)-1\left\}\right\{2(k+1)+1\}$
$=\frac{k(4k^2+6k-1)}{3}+(2k+2-1)(2k+2+1)$ [Using $\left(i\right)$]
$=\frac{k(4k^2+6k-1)}{3}+(2k+1)(2k+3)$
$=\frac{k(4k^2+6k-1)}{3}+(4k^2+8k+3)$
$=\frac{k(4k^2+6k-1)+3(4k^2+8k+3)}{3}$
$=\frac{4k^3+6k^2-k+12k^2+24k+9}{3}$
$=\frac{4k^3+18k^2+23k+9}{3}$
$=\frac{4k^3+14k^2+9k+4k^2+14k+9}{3}$
$=\frac{(k+1)(4k^2+14k+9)}{3}$
$=\frac{(k+1)\{4k^2+8k+4+6k+6-1\}}{3}$
$=\frac{(k+1)\left\{4\right(k^2+2k+1)+6(k+1)-1\}}{3}$
$=\frac{(k+1)\{4{(k+1)}^2+6(k+1)-1\}}{3}$
Thus, $P(k+1)$ is true whenever $P\left(k\right)$ is true.
Hence, by the principle of mathematical induction, statement $P\left(n\right)$ is true for all natural numbers i.e., $n$