Originally Posted by

**sjenkins** Ok, so I'm trying to use mathematical induction to prove 1^3+2^3+...+n^3=[n(n+1)/2]^2 whenever n is a positive integer. I think I have it, but I'm not sure, can any of you help me to make sure I'm on the right track? Thanks!

Basis step: P(1) states that the sum of the first one integer to the third power is [n(n+1)/2]^2. This is true since the sum of the first one integer to the third power is 1. [1(1+1)/2]^2= [1(2)/2]^2=(2/2)^2=1

Induction Step: Assume P(n) is true for a positive integer n that is

[2(2+1)/2]^2 [3(3+1)/2]^2

[2(3)/2]^2 [3(4)/2]^2

(6/2)^2 (12/2)^2

3^2 6^2

9=1^3+2^3 36=1^3+2^3+3^3

9=1+8 =9 36=1+8+27=36

Conclusion: The sum of the first n positive integers to the third power is given by [n(n+1)/2]^2 for all integers n>=1