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Closed form solution for series

  1. Mar 6, 2006 #1
    I am kind of blocked on how to find closed form solution for these finite series:

    sigma [nan]
    (n = 0 to k)
    k = any integer

    thanks for any hints.... and i can't find LaTex tutorials, where are they?
     
  2. jcsd
  3. Mar 6, 2006 #2

    benorin

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    Homework Helper

    So you may not know how to sum

    [tex]\sum_{n=0}^{k} na^n[/tex]

    so what. You do know how to sum something similar, right? like, say,

    [tex]\sum_{n=0}^{k} a^n[/tex]

    and is not [tex]\frac{d}{da}a^n = na^{n-1}[/tex]?

    so figure it out from there. (left click one of the expressions to see how to write them with symbols.)
     
    Last edited: Mar 7, 2006
  4. Mar 7, 2006 #3

    benorin

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    Homework Helper

    A new sum from an old sum

    To sum [tex]\sum_{n=0}^{k} na^n,[/tex]

    recall the sum of something similar, like, say,

    [tex]\sum_{n=0}^{k} a^n = \frac{1-a^{k+1}}{1-a}[/tex]

    and consder that [tex]\frac{d}{da}a^n = na^{n-1}[/tex]

    that one may write

    [tex] \frac{d}{da}\left( \sum_{n=0}^{k} a^n\right) = \frac{d}{da}\left( \frac{1-a^{k+1}}{1-a}\right)\Rightarrow \sum_{n=0}^{k}\left( \frac{d}{da}a^n\right) = \sum_{n=1}^{k}na^{n-1} =\frac{-(k+1)a^{k}(1-a)-(1-a^{k+1})(-1)}{(1-a)^2} [/tex]

    from the last equality, we have

    [tex]a\sum_{n=1}^{k}na^{n-1} =a\frac{-(k+1)a^{k}(1-a)+(1-a^{k+1})}{(1-a)^2}\Rightarrow \sum_{n=1}^{k}na^{n} =\frac{a}{1-a}\left[ \frac{1-a^{k+1}}{1-a}-(k+1)a^{k}\right] [/tex]

    and since [tex]0+\sum_{n=1}^{k}na^{n} = \sum_{n=0}^{k}na^{n}, [/tex]

    we have

    [tex]\boxed{ \sum_{n=0}^{k}na^{n} = \frac{a}{1-a}\left[ \frac{1-a^{k+1}}{1-a}-(k+1)a^{k}\right] = \frac{a}{(1-a)^2}\left[ 1+ka^{k+1}-(k+1)a^{k} \right] } [/tex]​
     
    Last edited: Mar 7, 2006
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