[krystar]

sigma[ 1^3+2^3+....+n^3 ] = (sigma [1+...+n] ) ^ 2?

i think iteration would prob be a good approach.

1^3=1^2
1^3 + 2^3 = (1+2) ^ 2
for n+1 iteration, you're adding [n+1]^3 to left. right side would be (1+2+....+n+[n+1])^2 for the equation to work.

so in order for it to be true, (1+2+....+n+[n+1])^2 must equal to (1+2+...+n)^2 + [n+1]^3

going to try to factor out the [n+1] out of the squared expression
(1+2+....+n+[n+1])^2 =
(1+2+...+n+[n+1])*(1+2+...+n+[n+1]) =
{using (a+b)*(c+d) = ac+bc+ad+bd rule}
(1+2+...+n)*(1+2+...+n) + [n+1](1+2+...+n) + (1+2+...+n)*[n+1] + [n+1]*[n+1] =
(1+2+...+n)^2 + 2*[n+1]*(1+2+...+n) + [n+1]^2

take step back. the first term is pulled out. (sigma(n))^2. so that means the rest must equal [n+1]^3

2*[n+1]*(1+2+...+n) + [n+1]^2 ?=? [n+1]^3
{expand right}
2*[n+1]*(1+2+...+n) + [n+1]^2 ?=? [n+1]^2*[n+1]
{expand right}
2*[n+1]*(1+2+...+n) + [n+1]^2 ?=? [n+1]^2*n + [n+1]^2
{subtract [n+1]^2 term both sides}
2*[n+1]*(1+2+...+n) ?=? [n+1]^2*n
{expand right}
2*[n+1]*(1+2+...+n) ?=? [n+1]*[n+1]*n
{elimninate [n+1] factor both sides}
2*(1+2+...+n) ?=? [n+1]*n


....and i'm stuck there. who wants to finish hehe