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Math Help - Sum the series

  1. #1
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    Sum the series

    Sorry, don't know how to use latex at the moment, but:

    Summation from k=0 to infinity of (k+2)2^(-k)

    \sum (k+2)2^{-k}

    It says to sum the series.

    I'm really unsure where to start in this problem. The only similarity I can notice is 1/(1-x) which equals the summation of x^k, but I don't think you can plug in -k's to change summations, can you?

    Any ideas?

    Edit-
    Nvm, put it in LaTex too.
    Last edited by ElTaco; December 15th 2010 at 06:18 AM. Reason: LaTex
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  2. #2
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    \displaystyle \sum_{k\ge 0}\frac{k+2}{2^k} = \sum_{k\ge 0}\frac{k}{2^k}+\sum_{k\ge 0}\frac{2}{2^k}. Putting \displaystyle x = \frac{1}{2} in \displaystyle \frac{2}{1-x} = 2\sum_{k\ge 0}x^k, gives  \displaystyle \sum_{k\ge 0}\frac{2}{2^k} = 4.
    Putting x = \frac{1}{2} in \displaystyle \frac{x}{(x-1)^2} = \sum_{k \ge 0}kx^k gives \displaystyle \sum_{k \ge 0}\frac{k}{2^k} = 2. Therefore \displaystyle \sum_{k\ge 0}\frac{k+2}{2^k}  = 6.
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  3. #3
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by ElTaco View Post
    Sorry, don't know how to use latex at the moment, but:

    Summation from k=0 to infinity of (k+2)2^(-k)

    \sum (k+2)2^{-k}

    It says to sum the series.

    I'm really unsure where to start in this problem. The only similarity I can notice is 1/(1-x) which equals the summation of x^k, but I don't think you can plug in -k's to change summations, can you?

    Any ideas?

    Edit-
    Nvm, put it in LaTex too.
    Is...

    \displaystyle \sum_{k=0}^{\infty} k\ x^{k} = x\ \sum_{k=1}^{\infty} k\ x^{k-1} = x\ \frac{d}{dx} \sum_{k=0}^{\infty} x^{k}=

    \displaystyle = x\ \frac{d}{dx} \frac{1}{1-x} = \frac{x}{(1-x)^{2}} (1)

    ... and...

    \displaystyle 2\ \sum_{k=0}^{\infty} (\frac{x}{2})^{k} = \frac{4}{2-x} (2)

    ... so that the sum of the series is the sum of (1) for x=\frac{1}{2} and (2) for x=1...

    \displaystyle \sum_{k=0}^{\infty} \frac{k+2}{2^{k}} = 2+4=6 (3)




    Merry Christmas from Italy

    \chi \sigma
    Last edited by chisigma; December 15th 2010 at 07:37 AM. Reason: corrected error in (1) [see Coffemachine's post]...
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