Question

Prove that $1^2+2^2+...+n^2>\frac{n^3}{3},nϵN$

Answer 1

$P\left(n\right):1^2+2^2+...+n^2>\frac{n^k}{3}$

$P\left(1\right):1>\frac{1^k}{3}$

Clearly for $k=1,1>\frac{1}{3}$

For $k=2,P\left(2\right):1+2^2>\frac{2^k}{3}$

$\Rightarrow 5>\frac{2^k}{3}$

For $k=1,P\left(2\right):5>\frac{2}{3}$ for $k=1,P\left(3\right):1+4+9>\frac{3}{3}$

For $k=2,P\left(2\right):5>\frac{4}{3}$ for $k=2,P\left(3\right):14>\frac{3^2}{3}$

For $k=3,P\left(2\right):5>\frac{8}{3}$ for $k=3,P\left(3\right):14>\frac{3^3}{3}$

For $k=4,P\left(2\right):5\ngtr \frac{16}{3}$ for $k=4,P\left(3\right):14\ngtr \frac{3^4}{3}$

$P\left(2\right)$ is satisfied for $k=1,2,3$

$\therefore k=3$, where $kϵZ^{+}$

Now we want to verify the value for $k\forall$ positive integral values of n using principle of mathematical induction

Verification:Let $P\left(m\right)$ be true for $n=m$ then