Question

$1^2+2^2+\dots +n^2>\frac{n^3}{3},n\in N$

Answer 1

Given that expression,

$1^2+2^2+.......n^2>\frac{n^3}{3},n\in N$

We know that sum of square of n natural numbers,

$1^2+2^2+.......n^2=\sum _{k=1}^n{k}^2=\frac{n(n+1)(2n+1)}{6}$

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

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

$(n+1)(2n+1)>n^2$

$n^2+3n+1>n^2$

$3n+1>0$

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

Hence, this is the answer.