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Math Help - Manipulating Sums

  1. #1
    Newbie evanator's Avatar
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    Cool Manipulating Sums

    Hi there,

    I am trying to solve the following problem. I have gotten a fair way into it, but the last steps elude me. Here is the problem:

    Show that \displaystyle\sum_{r=0}^n ^nC_r 2^r = 3^n

    So, we already know that \displaystyle\sum_{r=0}^n ^nC_r = 2^n, because

    (a+b)^n = \displaystyle\sum_{r=0}^n ^nC_r a^{n-r} b^{r}

    Letting a = 1 and b = 1:

    (1+1)^n = \displaystyle\sum_{r=0}^n ^nC_r 1^{n-r} 1^{r}

    2n = \displaystyle\sum_{r=0}^n ^nC_r

    Getting back to the matter at hand, we can take out the 2^n:

    2^n \displaystyle\sum_{r=0}^n 2^r

    We can also write it as:

    \displaystyle\sum_{r=0}^n 2^{n+r}

    This gives the same series, because

    (2^n \times 2^0) + (2^n \times 2^1) + (2^n \times 2^2) + ... + (2^n \times 2^n)

    is the same as

    2^{n+0} + 2^{n +1} + 2^{n+2} + ... + 2^{2n}

    I don't know what to do to get the 3^n though.
    Any help would be appreciated.

    Regards,

    Evanator
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  2. #2
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    Notice that 3^n=(1+2)^n. Just expand.
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  3. #3
    Newbie evanator's Avatar
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    Looks like I couldn't see the wood for the trees. Thank you, Plato.
    Expanding out, bearing in mind that (1+2)^n = 3^n:

    (a+b)^n = \displaystyle\sum_{r=0}^n ^nC_r a^{n-r} b^r

    (1+2)^n = \displaystyle\sum_{r=0}^n ^nC_r 1^{n-r} 2^r

    3^n = \displaystyle\sum_{r=0}^n ^nC_r 2^r
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