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Math Help - a^n - b^n

  1. #1
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    a^n - b^n

    Hi im wondering how one can obtain the formula a^n - b^n = (a-b) \sum_{k=0}^{n-1}a^{n-1-k}b^k out of the fact that (1-x)\sum_{k=0}^{n-1}x^{k}= 1 - x^n.

    I started via a^n - b^n = 1 -b^n -(1 - a^n) = (1-b)\sum_{k=0}^{n-1}b^{k} - (1-a)\sum_{k=0}^{n-1}a^{k} but don't know how to proceed. Any suggestions?
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  2. #2
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    Re: a^n - b^n

    Quote Originally Posted by EinStone View Post
    Hi im wondering how one can obtain the formula a^n - b^n = (a-b) \sum_{k=0}^{n-1}a^{n-1-k}b^k out of the fact that (1-x)\sum_{k=0}^{n-1}x^{k}= 1 - x^n.
    Set x = b / a.
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  3. #3
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    Re: a^n - b^n

    Wow, that was easy, thanks a lot.
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