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

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

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

    Letting $\displaystyle a = 1$ and $\displaystyle b = 1$:

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

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

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

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

    We can also write it as:

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

    This gives the same series, because

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

    is the same as

    $\displaystyle 2^{n+0} + 2^{n +1} + 2^{n+2} + ... + 2^{2n}$

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

    Regards,

    Evanator
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  2. #2
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    Notice that $\displaystyle 3^n=(1+2)^n$. 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 $\displaystyle (1+2)^n = 3^n$:

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

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

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