Question

Prove that $1^2+2^2+3^2+....+n^2>\frac{n^2}{3}$ for all $n\in N$ using principle of mathematical induction.

Answer 1

Formula is given by,

$1^2+2^2+3^2+......+n^2=\frac{n(n+1)(2n+1)}{6}$

To prove: $1^2+2^2+3^2+......+n^2>\frac{n^2}{3}$

But, $1^2+2^2+3^2+......+n^2=\frac{n(n+1)(2n+1)}{6}$

$\Rightarrow$ To prove $\frac{n(n+1)(2n+1)}{6}>\frac{n^2}{3}$

If we put $n=1$ , we get,

$\frac{1(1+1)(2+1)}{6}>\frac{1^2}{3}$

$1>\frac{1}{3}$

$\Rightarrow 1^2+2^2+3^2+......+n^2>\frac{n^2}{3}$