Question

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

Answer 1

Step $\left(1\right):$ Assume given statement
Let the statement $P\left(n\right)$ be defined as.
$P\left(n\right):1^2+2^2+\cdots +n^2>\frac{n^3}{3},n\in N$

Step $\left(2\right):$ Checking statement $P\left(n\right)$ for $n=1$
Put $n=1$ in $P\left(n\right)$, we get
$P\left(1\right):1^2>\frac{1^3}{3}$
$=1>\frac{1^3}{3}$ (Which is true)
Thus, $P\left(n\right)$ is true for $n=1$.

Step $\left(3\right):$ $P\left(n\right)$ for $n=K$
Put $n=K$ in $P\left(n\right)$ and assume that $P\left(K\right)$ is true for some natural number $K$ i.e.,
$P\left(K\right):1^2+2^2+\cdots +K^2>\frac{K^3}{3}\cdots \left(1\right)$

Step $\left(4\right):$ Checking statement $P\left(n\right)$ for $n=K+1$
Now, we shall prove that $P(K+1)$ is true whenever $P\left(K\right)$ is true.
We have,
$1^2+2^2+3^2+\cdots +K^2+(K+1{)}^2$
$(1^2+2^2+\cdots +K^2)+(K+1{)}^2$
$>\frac{K^3}{3}+(K+1{)}^2$
$>\frac{1}{3}[K^3+3(K^2+2k+1\left)\right]$
$>\frac{1}{3}[K^3+3K^2+6k+3]$
$>\frac{1}{3}\left[\right(K+1{)}^3+3K+2]$
$>\frac{1}{3}(K+1{)}^3$
Therefore, $P(K+1)$ is true whenever $P\left(K\right)$ is true.

Final Answer:
Hence, by principle of mathematical induction, $P\left(n\right)$ is true for all $nϵN$.