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Math Help - Question about summation

  1. #1
    Junior Member
    Joined
    Oct 2009
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    30

    Question about summation

    Hi, i'm not sure on how to do the summation of the following :

    n
    \sum (1-\frac{i-1}{n}).\frac{1}{n}
    i=1

    n
    \sum (1-\frac{i}{n}).\frac{1}{n}
    i=1

    The first one i got the answer is 1/2 (1 - 1/n) after simplifying everything.
    I'm stuck with the second one , not sure on how to sum up everything.
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  2. #2
    Junior Member
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    Oct 2009
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    \sum_{i\,=\,1}^n\left(1-\frac{i-1}n\right)\frac1n

    =\ \frac1n\sum_{i\,=\,1}^n1-\frac1{n^2}\sum_{i\,=\,1}^n(i-1)

    =\ \frac1n(n)-\frac1{n^2}\cdot\frac{n(n-1)}2

    =\ 1-\frac{n-1}{2n}

    =\ \frac{n+1}{2n}


    The second one is

    \sum_{i\,=\,1}^n\left(1-\frac in\right)\frac1n

    =\ \sum_{i\,=\,2}^{n+1}\left(1-\frac{i-1}n\right)\frac1n

    =\ \sum_{i\,=\,1}^{n}\left(1-\frac{i-1}n\right)\frac1n\,+\left(1-\frac {(n+1)-1}n\right)\frac1n-\left(1-\frac{1-1}n\right)\frac1n

    =\ \frac{n+1}{2n}+0-\frac1n

    =\ \frac{n-1}{2n}
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  3. #3
    Banned
    Joined
    Oct 2009
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    4,261
    Quote Originally Posted by xcluded View Post
    Hi, i'm not sure on how to do the summation of the following :

    n
    \sum (1-\frac{i-1}{n}).\frac{1}{n}
    i=1

    n
    \sum (1-\frac{i}{n}).\frac{1}{n}
    i=1

    The first one i got the answer is 1/2 (1 - 1/n) after simplifying everything.
    I'm stuck with the second one , not sure on how to sum up everything.

    \sum_{i=1}^n\left(1-\frac{i}{n}\right)\frac{1}{n}=\frac{1}{n}\left[ \sum_{i=1}^n\,1-\frac{1}{n}\sum_{i=1}^n\,i\right]=\frac{1}{n}\left[n-\frac{1}{n}\frac{n(n+1)}{2}\right]=....

    Tonio
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  4. #4
    Junior Member
    Joined
    Oct 2009
    Posts
    30
    Thanks guys !
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