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Math Help - Odd summs give cubes

  1. #1
    Junior Member theprof's Avatar
    Joined
    Apr 2005
    From
    Torino - Italy
    Posts
    26

    Odd sums give cubes

    Look at this

    1 = 1 = 1^3
    3 + 5 = 8 = 2^3
    7 + 9 + 11 = 27 = 3^3
    13 + 15 + 17 + 19 = 64 = 4^3
    ...

    nice, isnt'it?
    The proof is not that easy

    Bye
    Last edited by theprof; April 15th 2005 at 08:10 PM.
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  2. #2
    stringman
    Guest

    a simple solution

    The solution is easy.

    The sum is always of the type of q odd integer starting from 2 p+1 ending with 2(p+q)+1

    S_p,q=\Sum_k=p^p+q-1 (2k+1) = q (2p +q)

    At the first set we sum 1 term so q=1 and p=0, at the second step we sum 2 terms so q=2 and p=1 etc

    We just have to find the starting integer p at step q.


    But at set q we have skipped 1 + 2 + ... + q-1 = q(q-1)/2 odd integers therefore

    p = q (q-1)/2

    With this value of p

    S_p,q = q^3

    QED
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