Question

Prove by Mathematical induction that
$1^2+3^2+5^2...{(2n-1)}^2=\frac{n(2n-1)(2n+1)}{3}\forall n\in N$

Answer 1

TO PROVE:

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

PROOF:

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

$P\left(1\right):{(2\times 1-1)}^2=\frac{1(2-1)(2+1)}{3}$

$\Rightarrow {\left(1\right)}^2=1=\frac{1\times 1\times 3}{3}=1$

$\therefore$ L.H.S=R.H.S (Proved)

$\therefore P\left(1\right)$ is true.

Now, let $P\left(m\right)$ is true.

Then, ${P\left(m\right)=1}^2+3^2+5^2...+{(2m-1)}^2=\frac{m(2m-1)(2m+1)}{3}$

Now, we have to prove that $P(m+1)$ is also true.

${P(m+1)=1}^2+3^2+5^2...+{(2m-1)}^2+{[2(m+1)-1]}^2$

$=P\left(m\right)+{(2m+2-1)}^2$

$=P\left(m\right)+{(2m+1)}^2$

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

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

$=\frac{(2m+1)[m(2m-1)+3(2m+1)]}{3}$

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

$=\frac{(2m+1)[2m^2+5m+3]}{3}$

$=\frac{(2m+1)[2m^2+2m+3m+3]}{3}$

$=\frac{(2m+1)[2m(m+1)+3(m+1)]}{3}$

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

$=\frac{(m+1)[2(m+1)+1][2(m+1)-1]}{3}$

$\therefore p(m+1)$ is also true (Proved)