Question

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

Answer 1

Let p(n)

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

for n=1

LHS=$1^2=1$

RHS=$\frac{1\left(1\right)\left(3\right)}{3}=1$

so, p(n) is true for n=1

Let p(k) be true

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

for p (k + 1)

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

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

Form (1)

adding $(2k+1{)}^2$ on both sides.

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

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

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

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

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

so, using mathematical induction

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