Question

$1^2+3^2+5^{2}5+...+(2n-1{)}^2=\frac{1}{3}n(4n^2-1)$

Answer 1

Let P(n) : $1^2+3^2+5^2+.....+(2n-1{)}^2=\frac{1}{3}n(4n^2-1)$

For n = 1

$1=\frac{1}{3}.1.(4-1)$

$1=1$

$\Rightarrow$ P(n) is true for n = 1

Let P(n) is true for n = k, so

$1^2+3^2+5^2+......+(2k-1{)}^2$

$=\frac{1}{3}k(4k^2-1)$ .......(1)

We have to show that,

$1^2+3^2+5^2+....+(2k-1{)}^2+(2k+1{)}^2=\frac{1}{3}(k+1)\left[4\right(k+1{)}^2-1]$

Now,

$\{1^2+3^2+5^2+...+(2k-1{)}^2\}+(2k+1{)}^2$

$=\frac{1}{3}k(4k^2-1)+(2k+1{)}^2$

[Using equation (1)]

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

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

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

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

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

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

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

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

$=\frac{(k+1)}{2}[4k^2+6k+2k+3]$

$=\frac{(k+1)}{2}[4k^2+8k+4-1]$

$=\frac{(k+1)}{2}\left[4\right(k+1{)}^2-1]$

$\Rightarrow$ P(n) is true for n = k + 1

$\Rightarrow$ P(n) is true for all n $ϵ$ N by PMI