Question

1square+2square+......+nsquare>ncube÷3show by using mathematical induction

Answer 1

We have to prove by mathematical induction that,

$1^2+2^2+3^2+\dots\ +n^2>\dfrac{n^3}{3},\ \ n\in\mathbb{N}$

Let n = 1.

$\text{LHS = 1^2=1 }\\ \\ \text{RHS = \dfrac{1^3}{3}=\dfrac{1}{3} }\\ \\ \\ \text{Here, 1>\dfrac{1}{3} }$

Let n = k. Assume that,

$1^2+2^2+3^2+\dots\ +k^2>\dfrac{k^3}{3},\ \ k\in\mathbb{N}$

Let  $1^2+2^2+3^2+\dots\ +k^2=\dfrac{k^3}{3}+m,\quad m\in\mathbb{N}$

Let n = k + 1.

$\begin{aligned}&1^2+2^2+3^2+\dots\ +k^2+(k+1)^2\\ \\ \implies\ \ &\dfrac{k^3}{3}+m+k^2+2k+1\\ \\ \implies\ \ &\dfrac{k^3+3k^2+6k+3}{3}+m\\ \\ \implies\ \ &\dfrac{(k+1)^3}{3}+\left(\dfrac{3k+2}{3}+m\right)\\\\\implies\ \ &1^2+2^2+3^2+\dots\ +(k+1)^2>\dfrac{(k+1)^3}{3}\end{aligned}$

Hence Proved!