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Math Help - summation

  1. #1
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    summation

    Justify:

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

    <br />
\sum_{i=n}^{2n} i = \frac{3n(n+1)}{2}<br />
    Since
    <br />
\sum_{i=1}^{n} i = \frac{n(n+1)}{2}<br />
and <br />
\sum_{i=1}^{2n} i = \frac{2n(2n+1)}{2}<br />

    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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    \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
    <br />
\sum_{i=1}^{n} i = \frac{n(n+1)}{2}<br />
and <br />
\sum_{i=1}^{2n} i = \frac{2n(2n+1)}{2}<br />

    If you subtracted it wouldn't you get:

    <br />
\frac{2n(2n+1)}{2} - \frac{n(n+1)}{2}<br />

    <br />
\frac{4n^2 + 2n - n^2 - n}{2}<br />

    <br />
\frac{3n^2 + n}{2}<br />

    <br />
\frac{n(3n+1)}{2}<br />

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

    \sum\limits_{k = 1}^{2n} k  = \frac{{2n\left( {2n + 1} \right)}}<br />
{2}

    \sum\limits_{k = 1}^{n - 1} k  = \frac{{\left( {n - 1} \right)n}}<br />
{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:

    <br />
\frac{2n(2n+1)}{2} - \frac{n(n+1)}{2}<br />

    <br />
\frac{4n^2 + 2n - n^2 - n}{2}<br />

    <br />
\frac{3n^2 + n}{2}<br />

    <br />
\frac{n(3n+1)}{2}<br />

    That's not what it says : (
    yes but

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

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