# Prove that  $1^2+2^2+...+n^2 > \large\frac{n^3}{3}$$, n \in N ## 1 Answer Let P(n) be the given statement i.e., P(n) : 1^2+2^2+...+n^2 > \large\frac{n^3}{3}$$, n \in N$
We note that $P(n)$ is true for $n=1$ since $1^2 > \large\frac{1^3}{3}$
Assume that $P(k)$ is true.
i.e., $\qquad P(k) : 1^2+2^2+...+k^2 > \large\frac{k^3}{3}$----------(1)
We shall now prove that $P(k+1)$ is true whenever $P(k)$ is true.
We have $1^2+2^2+3^2+...+k^2+(k+1)^2$
$\qquad = (1^2+2^2+...+k^2)+(k+1)^2> \large\frac{k^3}{3}$$+(k+1)^2 \qquad [ by (1) ] \qquad = \large\frac{1}{3}$$ [ k^3+3k^2+6k+3]$
$\qquad = \large\frac{1}{3}$$[ (k+1)^3+3k+2]> \large\frac{1}{3}$$(k+1)^3$
Therefore, $P(k+1)$ is also true whenever $P(k)$ is true. Hence, by mathematical induction $P(n)$ is true for all $n \in N$