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Math Help - Not sure how to start this

  1. #1
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    Not sure how to start this

    Prove that the sum of the first n cubes is given by

    1^3+2^3+3^3+4^3+.....+n^3=(n^2(n+1)^2)/4
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  2. #2
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    Quote Originally Posted by padsinseven View Post
    Prove that the sum of the first n cubes is given by
    1^3+2^3+3^3+4^3+.....+n^3=(n^2(n+1)^2)/4
    INDUCTION?
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  3. #3
    Super Member wingless's Avatar
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    You can prove this by using Induction.
    There are two steps:
    1) Try your equation with a positive integer. If its equal, get to the second step.
    2) Show that its equal for f(n) and f(n+1)

    1^3 + 2^3 + 3^3 + ........ + n^3 = (\frac {n(n+1)}{2})^2

    1) 1^3 + 2^3 + 3^3 = (\frac {3(3+1)}{2})^2 = 36

    2) f(n) = 1^3 + 2^3 + 3^3 + ........ + n^3 = (\frac {n(n+1)}{2})^2
    f(n+1) = 1^3 + 2^3 + 3^3 + ........ + n^3 + (n+1)^3 = (\frac {(n+1)(n+2)}{2})^2
    If you noticed, f(n+1) includes f(n). So we will replace 1^3 + 2^3 + 3^3 ....... + n^3 with (\frac {n(n+1)}{2})^2

    f(n+1) = (\frac {n(n+1)}{2})^2 + (n+1)^3 = (\frac {(n+1)(n+2)}{2})^2
    0 = 0, it means this equation is true.

    What did we do there?
    First we tried our formula for an integer and saw that it works.
    Then, we proved that it'll work for every f(n+1). Which means its true for every n.
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