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Thread: Is is possible to use Arithmetic progression in this case?

  1. #1
    Junior Member
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    Is is possible to use Arithmetic progression in this case?

    $\displaystyle n$
    $\displaystyle \sum e^\frac{i}{n}(\frac{1}{n})$
    $\displaystyle i = 1$

    $\displaystyle \frac{1}{n}[ e^\frac{1}{n} + e^\frac{2}{n} + e^\frac{3}{n} +.....+ e^\frac{n}{n} ]$

    Did i compute wrongly ? I find it a bit weird. I can't proceed further on combining the bracket terms.

    Thanks.
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  2. #2
    Member
    Joined
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    New Delhi
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    Quote Originally Posted by xcluded View Post
    $\displaystyle n$
    $\displaystyle \sum e^\frac{i}{n}(\frac{1}{n})$
    $\displaystyle i = 1$

    $\displaystyle \frac{1}{n}[ e^\frac{1}{n} + e^\frac{2}{n} + e^\frac{3}{n} +.....+ e^\frac{n}{n} ]$

    Did i compute wrongly ? I find it a bit weird. I can't proceed further on combining the bracket terms.

    Thanks.
    $\displaystyle [ e^\frac{1}{n} + e^\frac{2}{n} + e^\frac{3}{n} +.....+ e^\frac{n}{n} ]$
    it is in geometric progression with common ratio
    $\displaystyle = e^\frac{1}{n}$
    use sum of infinite G.P. terms formula
    $\displaystyle = \frac{a}{1-r}$
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