See title

  • @NeoNachtwaechter
    8•18 hours ago

    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?

    • @DreamlandLividity
      5•15 hours ago
      Proof by induction?

      1±2±3±...±n =(1+n)*n/2

      plugging 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)/4

      If this holds for n: 1^3 + 2^3 +3^3 + ... + n^3 = (n^4 + 2n^3 +n^2)/4

      Then 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 + 1

      n^3 + 2n^2 + n + n^2 + 2n + 1 =? n^3 + 3n^2 + 3n + 1

      n^3 + 3n^2 + 3n + 1 =? n^3 + 3n^2 + 3n + 1

      Which is obviously true.

      So yes, it holds forever.

    • @[email protected]
      2•17 hours ago
      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.

      • @NeoNachtwaechter
        2•16 hours ago

        You are right LOL: I do not approve. But somehow I like the lazy approach :)