See title
Look at these equations:
1^3 = 1^2
1^3 + 2^3 = (1+2)^2
1^3 + 2^3 +3^3 = (1+2+3)^2
1^3 + 2^3 +3^3 +4^3 = (1+2+3+4)^2
Question:
Can it go on like this forever, is it always a true equation? If yes, why? If no, why?
Your math teacher might not approve of this proof
The given examples suffice to prove the general identity. Both sides are obviously degree 4 polynomials, so if they agree at 5 points (include the degenerate case 0^3 = 0^2), then they agree everywhere.
You are right LOL: I do not approve. But somehow I like the lazy approach :)
Corresponding Wikipedia page with a graphical proof, among others
The graphical proof is really nice :)
Proof by induction?
1±2±3±...±n =(1+n)*n/2plugging that into the right side of the equation to transform it:
((1+n)*n/2)^2 = (1+n)^2*n^2/4=n^2(n^2+2n+1)/4 = (n^4 + 2n^3 +n^2)/4If this holds for n:
1^3 + 2^3 +3^3 + ... + n^3 = (n^4 + 2n^3 +n^2)/4Then for n+1:
(n^4 + 2n^3 +n^2)/4 + (n+1)^3 =? (1+n + 1)^2*(n+1)^2/4(n^4 + 2n^3 +n^2)/4 + (n+1)^3 =? (n^2+4n + 4)(n^2 +2n + 1)/4(n^4 + 2n^3 +n^2)/4 + (n+1)^3 =? (n^4 + 4n^3 + 4n^2 + 2n^3 + 8n^2 + 8n + n^2 + 4n + 4)/4(n^4 + 2n^3 +n^2)/4 + (n+1)^3 =? (n^4 + 2n^3 + n^2)/4 + (4n^3 + 12n^2 + 12n + 4)/4(n+1)(n^2 +2n + 1) =? n^3 + 3n^2 + 3n + 1n^3 + 2n^2 + n + n^2 + 2n + 1 =? n^3 + 3n^2 + 3n + 1n^3 + 3n^2 + 3n + 1 =? n^3 + 3n^2 + 3n + 1Which is obviously true.
So yes, it holds forever.
This is the way.