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Math Help - Sum of the series

  1. #1
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    Sum of the series

    Not sure where to go with this:
     <br />
\sum{3^n\over(1-2x)^n*n!}<br />

    I tried breaking it up into:
    \sum{1\over(1-2x)^n} + \sum{3^n\over n!}

    Then
    \sum{3^n\over n!} would become e^3-1

    And now I'm stuck.
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  2. #2
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    Quote Originally Posted by Capnredbeard View Post
    Not sure where to go with this:
     <br />
\sum{3^n\over(1-2x)^n*n!}<br />

    I tried breaking it up into:
    \sum{1\over(1-2x)^n} + \sum{3^n\over n!}

    Then
    \sum{3^n\over n!} would become e^3-1

    And now I'm stuck.

    How did you break up that summation?
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  3. #3
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    You're right. I can't break it up like that, but I can do this.
    \sum{{{3\over1-2x}^n}*{1\over n!}}

    And {1\over n!} = e
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  4. #4
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    Is there any indexing for this series?
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  5. #5
    Super Member Failure's Avatar
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    Quote Originally Posted by Capnredbeard View Post
    Not sure where to go with this:
     <br />
\sum{3^n\over(1-2x)^n*n!}<br />
    If the summation is indexed by n then you have simply

    \sum_{n=0}^\infty \frac{3^n}{(1-2x)^n n!}=\sum_{n=0}^\infty \frac{1}{n!}\left(\frac{3}{1-2x}\right)^n=\mathrm{e}^{\frac{3}{1-2x}}
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  6. #6
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    n=0 to infinity
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