Question

By the principle of Mathematical induction, prove that, for $n\ge 1.$
$1^2+2^2+3^2+\cdots +n^2>\frac{n^3}{3}.$

Answer 1

Let $P\left(n\right)$ be the given statement, i.e, $P\left(n\right):1^2+2^2+.......+n^2>\frac{n^3}{3},n\in N$

for $n=1$ $\because 1^2>\frac{1^3}{3}$ is true.

Assure that $P\left(k\right)$ is true i.e, $P\left(k\right):1^2+2^2+......+k^2>\frac{k^3}{3}⟶\left(1\right)$

Now we will prove that $P(k+1)$ is true when $P\left(k\right)$ is true.

we have, $1^2+2^2+......+k^2+{(k+1)}^2$

$=(1^2+2^2+........+k^2)+{(k+1)}^2$

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

$=1[k^3+3(k^2+2k+1)]$

$=\frac{1}{3}[k^3+3k^2+6k+3]$

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

$>\frac{1}{3}{(k+1)}^3.$

$\Rightarrow 1^2+2^2+3^2+..........+k^2+{(k+1)}^2>\frac{1}{3}{(k+1)}^3$

Hence, proved.