Question

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

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 $k=2,P\left(2\right):1+2^2>\frac{2^k}{3}⟹5>\frac{2^k}{3}$For $k=1,P\left(2\right):5>\frac{2}{3}$For $k=2,P\left(2\right):5>\frac{4}{3}$For $k=3,P\left(2\right):5>\frac{8}{3}$For $k=4,P\left(2\right):5\ngtr \frac{16}{3}$For $k=1,P\left(3\right):1+4+9>\frac{3}{3}$For $k=2,P\left(3\right):14>\frac{9}{3}$For $k=3,P\left(3\right):14>\frac{27}{3}$For $k=4,P\left(4\right):5>\frac{3^4}{3}$$P\left(2\right)$ is satisfied for $k=1,2,3$$\therefore k=3$, where $k\in Z^{+}$Now we want to verify the value of $k\forall$ positive integral values of $n$ using principle of mathematical induction.Verification: Let $P\left(m\right)$be true for $n=m$, then$P\left(m\right):1^2+2^2+...+m^2>\frac{m^3}{3}$We shall now prove that $P(m+1)$ is true, i.e.,$1^2+2^2+...+m^2+(m+1{)}^2>\frac{(m+1{)}^3}{3}⟹P(m+1)$ is trueNow $P\left(m\right)$ is true.$1^2+2^2+...+m^2+(m+1{)}^2>\frac{(m{)}^3}{3}+(m+1{)}^2⟹1^2+2^2+...+m^2+(m+1{)}^2>\frac{1}{3}(m^3+3m^2+6m+3)⟹1^2+2^2+...+m^2+(m+1{)}^2>\frac{1}{3}\left[\right(m^3+3m^2+3m+1)+(3m+2\left)\right]⟹1^2+2^2+...+m^2+(m+1{)}^2>\frac{1}{3}[(m+1{)}^3+(3m+2\left)\right]>\frac{(m+1{)}^3}{3}$Therefore, $P\left(m\right)$ is true $⟹P(m+1)$ is true