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Math Help - Can someone help me understand how to evaluate sums?

  1. #1
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    Can someone help me understand how to evaluate sums?

    Hello,
    I was just wondering if someone can help me understand how to go about evaluating sums???

    I am very new and lost as to how to get started... I don't understand the idea behind it....


    For example:
    2n
    E i
    i=5



    thanks in advance!
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  2. #2
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    \sum\limits_{i = 1}^{2n} i  = \frac{{2n\left( {2n + 1} \right)}}{2} = n\left( {2n + 1} \right)
    And
    \sum\limits_{i = 5}^{2n} i  = \sum\limits_{i = 1}^{2n} i  - \sum\limits_{i = 1}^4 i

    So?
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  3. #3
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    Thanks for the reply...

    I don't really understand what each step is suppose to accomplish but....

    But, trying to draw from your example, is this correct?
    [2n(2n+1)+2n(2n+2)+2n(2n+3)+2n(2n+4)+2n(2n+5)]/2 ?????????? and the result is 25??

    Thanks again!
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  4. #4
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    Quote Originally Posted by matthayzon89 View Post
    I don't really understand what each step is suppose to accomplish but....But, trying to draw from your example, is this correct?
    [2n(2n+1)+2n(2n+2)+2n(2n+3)+2n(2n+4)+2n(2n+5)]/2 ?????????? and the result is 25??
    The answer is \sum\limits_{i = 5}^{2n} i  = n\left( {2n + 1} \right) - 10 = 2n^2  + n - 10
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  5. #5
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    Where did the -10 come from?

    Since this is finite sum evaluation couldn't I evaluate it to a specific number?

    Thanks
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  6. #6
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    Quote Originally Posted by matthayzon89 View Post
    Where did the -10 come from?
     \sum\limits_{i = 1}^4 i  = 10
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  7. #7
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    Can someone help me understand what N is worth here? do I just count n as 1? or what?

    4n
    E(n^3)i
    i=0
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  8. #8
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    Quote Originally Posted by matthayzon89 View Post
    Can someone help me understand what N is worth here? do I just count n as 1? or what?
    4n
    E(n^3)i
    i=0
    I don't think you will understand this but here goes:
    \sum\limits_{i = 0}^{4n} {n^3 i}  = n^3 \sum\limits_{i = 0}^{4n} i  = n^3 \left( {\frac{{4n\left( {4n + 1} \right)}}<br />
{2}} \right) = n^3 \left( {2n(4n + 1)} \right)
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  9. #9
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    In the expression \sum\limits_{i = 0}^{4n} {n^3 i}, n is unknown. This is the sum n^3\cdot 0+n^3\cdot 1+n^3\cdot2+\dots+n^3\cdot4n with 4n+1 terms. Until the value of n is given, it cannot be evaluated to a number.
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