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Math Help - Anyone intrested in checking sigma notation?

  1. #1
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    Anyone intrested in checking sigma notation?

    Could someone please check my work?

    107
    Σ (4+3i)
    i=29

    107 107
    Σ (4+3i)=4*107+3Σi
    i=29 29

    =9737

    Many thanks
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  2. #2
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    This is what MathCad makes of your sum.
    Attached Thumbnails Attached Thumbnails Anyone intrested in checking sigma notation?-sum.gif  
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  3. #3
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    Quote Originally Posted by stewpot View Post
    Code:
    107
     Σ (4+3i)
    i=29
    
    107                  107
     Σ (4+3i) = 4*107 + 3 Σi
    i=29                 29
    I think you may have forgotten about your lower limit...?

    You are not summing from i = 1 to i = 107; you are summing from i = 29 to i = 107. As such, you should have only 107 - 29 + 1 = 81 terms. This makes the first summation that you broke out, the "four times the summation", equal to 4*81 = 324, not 4*107 = 428.

    The second summation can be represented as three times the difference of the sum of the first 107 numbers and the sum of the first 28 numbers.

    Apply the formula, do the subtraction, and multiply by 3. Then add in the "four times" part.
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  4. #4
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    Quote Originally Posted by stewpot View Post
    Could someone please check my work?

    107
    Σ (4+3i)
    i=29

    107 107
    Σ (4+3i)=4*107+3Σi
    i=29 29

    =9737

    Many thanks
    \sum_{i = 29}^{107}{4 + 3i}

    First, it would help if you figured out how many terms there will be. In this case there are 79.

    \sum_{i = 29}^{107}{4 + 3i} = (4 + 3\times 29) + (4 + 3\times 30) + (4 + 3\times 31) + \dots + (4 + 3 \times 107)

     = 79 \times 4 + 3\sum_{i = 29}^{107}i.

    This sum is an arithmetic series - so you solve it using the S_n = \frac{n}{2}(t_1 + t_n) formula.

     = 316 + 3\left[\frac{79}{2}(29 + 107)\right]

     = 316 + 3\times 5372

     = 316 + 16116

     = 16432.
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  5. #5
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    Sigma Notation

    Brill, many thanks
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