Question 393642
The base case {{{1^3 = 1^2}}} satisfies. Now, suppose that


{{{sum(i^3, i = 1, n) = (sum(i, i = 1, n))^2}}} for arbitrary n. We wish to prove that this implies the {{{n+1}}} case will be true, or that


{{{sum(i^3, i = 1, n+1) = (sum(i, i = 1, n+1))^2}}}


We can show this is true, using the fact that {{{sum(i^3, i = 1, n+1) = sum((i^3), i = 1, n) + (n+1)^3 = (sum(i, i = 1, n))^2 + (n+1)^3 = ((n(n+1))/2)^2 + (n+1)^3}}} and {{{(sum(i, i = 1, n+1))^2 = ((n+1)(n+2)/2)^2}}} Therefore this is equivalent to showing that


{{{((n(n+1))/2)^2 + (n+1)^3 = ((n+1)(n+2)/2)^2}}} 


Divide both sides by {{{(n+1)^2}}}


{{{n^2/4 + (n+1) = ((n+2)/2)^2}}}


{{{(n^2 + 4n + 4)/4 = (n^2 + 4n + 4)/4}}}


This holds for all n, so the statement holds for n+1. The induction is complete, so {{{sum(i^3, i = 1, n) = (sum(i, i = 1, n))^2}}}