Question

1²+2²+3²+........+n²>n³/3​

Answer 1

Can anyone prove this using mathematical induction:

1^3+2^3+…+n^3 = n(n+1)……..^2

……………………---------

………………………..2

for all integers n >=1.

Note n(n+1) / 2 are in brackets and than all that is squared by 2 (^2).

Answer 2

Suppose this proposition is true for a

particular n (and you can see that it is

certainly true for the first few positive

integers).

Then 1^2 + 2^2 + ... + n^2 + (n+1)^2

>n^3/3 + (n+1)^2

= (1/3)(n+1)^3 - n^2 - n - 1/3 + n^2 + 2n + 1

= (1/3)(n+1)^3 +n + 1

>(1/3)(n+1)^3.

This demonstrates that if the proposition is

true for "n"

it is also true for the successor of "n". Since

the proposition is indeed true for n=1, it will be

true for all n