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Thread: summation

  1. #1
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    summation

    Justify:

    $\displaystyle
    \sum_{i=n}^{2n} i = \frac{3n(n+1)}{2}
    $
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  2. #2
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    Recall the $\displaystyle \sum\limits_{k = 1}^J k = \frac{{J\left( {J + 1} \right)}}{2}$.
    Then note that $\displaystyle \sum\limits_{k = 1}^{2n} k = \sum\limits_{k = 1}^{n - 1} k + \sum\limits_{k = n}^{2n} k
    $
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  3. #3
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    Quote Originally Posted by qzno View Post
    Justify:

    $\displaystyle
    \sum_{i=n}^{2n} i = \frac{3n(n+1)}{2}
    $
    Since
    $\displaystyle
    \sum_{i=1}^{n} i = \frac{n(n+1)}{2}
    $ and $\displaystyle
    \sum_{i=1}^{2n} i = \frac{2n(2n+1)}{2}
    $

    so subtracting the first from the second and adding n gives your answer. (Too slow :-))
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  4. #4
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    $\displaystyle \sum\limits_{i=n}^{2n}{i}=\sum\limits_{i=0}^{n}{(i +n)}.$
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  5. #5
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    Quote Originally Posted by danny arrigo View Post
    Since
    $\displaystyle
    \sum_{i=1}^{n} i = \frac{n(n+1)}{2}
    $ and $\displaystyle
    \sum_{i=1}^{2n} i = \frac{2n(2n+1)}{2}
    $

    If you subtracted it wouldn't you get:

    $\displaystyle
    \frac{2n(2n+1)}{2} - \frac{n(n+1)}{2}
    $

    $\displaystyle
    \frac{4n^2 + 2n - n^2 - n}{2}
    $

    $\displaystyle
    \frac{3n^2 + n}{2}
    $

    $\displaystyle
    \frac{n(3n+1)}{2}
    $

    That's not what it says : (
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  6. #6
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    $\displaystyle \sum\limits_{k = 1}^{2n} k = \sum\limits_{k = 1}^{n - 1} k + \sum\limits_{k = n}^{2n} k $

    $\displaystyle \sum\limits_{k = 1}^{2n} k = \frac{{2n\left( {2n + 1} \right)}}
    {2}$

    $\displaystyle \sum\limits_{k = 1}^{n - 1} k = \frac{{\left( {n - 1} \right)n}}
    {2}$

    Now try it again.
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  7. #7
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    Quote Originally Posted by qzno View Post
    If you subtracted it wouldn't you get:

    $\displaystyle
    \frac{2n(2n+1)}{2} - \frac{n(n+1)}{2}
    $

    $\displaystyle
    \frac{4n^2 + 2n - n^2 - n}{2}
    $

    $\displaystyle
    \frac{3n^2 + n}{2}
    $

    $\displaystyle
    \frac{n(3n+1)}{2}
    $

    That's not what it says : (
    yes but

    Quote Originally Posted by danny arrigo View Post
    Since
    $\displaystyle
    \sum_{i=1}^{n} i = \frac{n(n+1)}{2}
    $ and $\displaystyle
    \sum_{i=1}^{2n} i = \frac{2n(2n+1)}{2}
    $

    so subtracting the first from the second and adding n gives your answer. (Too slow :-))
    We need the extra part - adding $\displaystyle n$
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