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Math Help - Sum to n of series

  1. #1
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    Sum to n of series

    If S_n=\displaystyle \sum^n_{r=0} a^r(1+a+a^2+...+a^r), (|a|\neq 1) by considering (1-a)S_n, show that
    S_n=\frac{1-a^{2n+2}}{(1-a^2)(1-a)}-\frac{a^{n+1}(1+a^{n+1})}{(1-a)^2}
    State the set of values of a for which S_napproaches a limit as n \rightarrow\infty and find the sum to infinity of the series for these values.

    Attempt:
    \displaystyle \sum^n_{r=0} a^r(1+a+a^2+...+a^r)
    =\displaystyle \sum^n_{r=0} a^r\left(\displaystyle \sum^n_{r=0}a^r\right)
    =\displaystyle \sum^n_{r=0}a^r \left(\frac{1-a^r}{1-a}\right)
    =\displaystyle \sum^n_{r=0}\frac{a^r-a^{2r}}{1-a}
    Am I correct so far? and how to continue because I don't see a way of summing that?
    Thanks!
    Last edited by arze; August 17th 2010 at 04:04 AM.
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  2. #2
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    This is almost correct. Here is the correct version:

    \displaystyle \sum^n_{r=0} a^r\left(\displaystyle \sum^r_{k=0}a^k\right)

    =\displaystyle \sum^n_{r=0}a^r \left(\frac{1-a^{r+1}}{1-a}\right)

    In the question you are told to consider (1-a)S_n. Why don't you try that!
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  3. #3
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    (1-a)S_n=S_n-aS_n
    =\displaystyle \sum^n_{r=0}a^r\left(\frac{(1-a^{r+1})(1-a)}{1-a}\right)
    =\displaystyle \sum^n_{r=0}a^r(1-a^{r+1})
    \displaystyle \sum^n_{r=0}a^r-\sum^n_{r=0}a^{2r+1}
    =\frac{1-a^{n+1}}{1-a}-\frac{a-a^{2n+2}}{1-a^2}

    S_n=\frac{1-a^{n+1}}{(1-a)^2}-\frac{a-a^{2n+2}}{(1-a^2)(1-a)}
    Which isn't the given one and I can't see where I've gone wrong
    Thanks!
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  4. #4
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    Quote Originally Posted by arze View Post
    If S_n=\displaystyle \sum^n_{r=0} a^r(1+a+a^2+...+a^r), (|a|\neq 1) by considering (1-a)S_n, show that
    S_n=\frac{1-a^{2n+2}}{(1-a)^2(1-a)}-\frac{a^{n+1}(1+a^{n+1})}{(1-a)^2}
    State the set of values of a for which S_napproaches a limit as n \rightarrow\infty and find the sum to infinity of the series for these values.

    Attempt:
    \displaystyle \sum^n_{r=0} a^r(1+a+a^2+...+a^r)
    =\displaystyle \sum^n_{r=0} a^r\left(\displaystyle \sum^n_{r=0}a^r\right)
    =\displaystyle \sum^n_{r=0}a^r \left(\frac{1-a^r}{1-a}\right)
    =\displaystyle \sum^n_{r=0}\frac{a^r-a^{2r}}{1-a}
    Am I correct so far? and how to continue because I don't see a way of summing that?
    Thanks!
    Dear arze,

    When n=1,

    S_1=\displaystyle\sum_{r=0}^{1}{a^r(1+a+a^2+.....+  a^r)}=a^0+a(1+a)

    But if you consider the expression, \frac{1-a^{2n+2}}{(1-a)^2(1-a)}-\frac{a^{n+1}(1+a^{n+1})}{(1-a)^2}......

    When n=1,

    \frac{1-a^{2n+2}}{(1-a)^2(1-a)}-\frac{a^{n+1}(1+a^{n+1})}{(1-a)^2}

    =\frac{1-a^{4}}{(1-a)^3}-\frac{a^{2}(1+a^{2})}{(1-a)^2}

    =\frac{1+a^2}{1-a^2}\left(\frac{1-a^2}{1-a}-a^2\right)

    =\frac{1+a^2}{1-a^2}\left(1+a-a^2\right)

    Therefore, S_1\neq\frac{1-a^{2n+2}}{(1-a)^2(1-a)}-\frac{a^{n+1}(1+a^{n+1})}{(1-a)^2}
    Last edited by Sudharaka; August 17th 2010 at 02:23 AM.
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  5. #5
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    Yea, that's not the correct identity. The correct identity is

    [tex]\displaystyle \displaystyle S_n = \frac{(1-a^{n+1})(1-a^{n+2})}{(1-a)^2(1+a)}[/Math]
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  6. #6
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    Quote Originally Posted by Vlasev View Post
    Yea, that's not the correct identity. The correct identity is

    [tex]\displaystyle \displaystyle S_n = \frac{(1-a^{n+1})(1-a^{n+2})}{(1-a)^2(1+a)}[/Math]
    But that doesn't give me the result I'm supposed to get either
    S_n=\frac{1-a^{2n+2}}{(1-a^2)(1-a)}-\frac{a^{n+1}(1+a^{n+1})}{(1-a)^2}
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  7. #7
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    Quote Originally Posted by arze View Post
    But that doesn't give me the result I'm supposed to get either
    S_n=\frac{1-a^{2n+2}}{(1-a^2)(1-a)}-\frac{a^{n+1}(1+a^{n+1})}{(1-a)^2}
    Because what you are asked to get is (in this form) false! Unless you typed something incorrectly in your question.
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  8. #8
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    hmmm, i checked the book again and i did not type anything wrong. Well, thank you very much!
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  9. #9
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    \displaystyle\ S_n=\sum_{r=0}^na^r\left(1+a+a^2+....+a^r\right)

    \displaystyle\ aS_n=\sum_{r=0}^na^{r+1}\left(1+a+a^2+....+a^r\rig  ht)

    \displaystyle\ (1-a)S_n=\sum_{r=0}^n\left(a^r+a^{r+1}+a^{r+2}+....+a  ^{2r}\right)-\sum_{r=0}^n\left(a^{r+1}+a^{r+2}+....+a^{2r+1}\ri  ght)

    \displaystyle\ =\sum_{r=0}^na^r-\sum_{r=0}^na^{2r+1}

    The first sum has n+1 terms, the first term is 1 and the common ratio is a.
    The second sum has (2n+2)/2 =n+1 terms, the first term is a and the common ratio is a^2.


    \displaystyle\ (1-a)S_n=\frac{1-a^{n+1}}{1-a}-a\left(\frac{1-a^{2(n+1)}}{1-a^2}\right)

    \displaystyle\ S_n=\frac{1-a^{n+1}}{(1-a)^2}-a\left(\frac{1-a^{2n+2}}{(1-a)^2(1+a)}\right)

    \displaystyle\huge\ =\frac{(1+a)\left(1-a^{n+1}\right)-a\left(1-a^{2n+2}\right)}{(1-a)^2(1+a)}

    \displaystyle\huge\ =\frac{(1+a)\left(1-a^{n+1}\right)-a\left(1-a^{n+1}\right)\left(1+a^{n+1}\right)}{(1-a)^2(1+a)}

    \displaystyle\huge\ =\frac{\left(1-a^{n+1}\right)\left[1+a-a\left(1+a^{n+1}\right)\right]}{(1-a)^2(1+a)}

    \displaystyle\huge\ =\frac{\left(1-a^{n+2}\right)\left(1-a^{n+1}\right)}{(1-a)^2(1+a)}

    so I'd have to agree with Vlasev.
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