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anonymous

  • one year ago

Does anyone know how to do this Sigma Notation problem?

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  1. anonymous
    • one year ago
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  2. freckles
    • one year ago
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    it starts at i=5 and ends at i=13 you do replace i with 5 + sign replace i with 6 + sign replace i with 7 + sign replace i with 8 + sign replace i with 9 + sign replace i with 10 + sign replace i with 11 + sign replace i with 12 + sign replace i with 13 end

  3. freckles
    • one year ago
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    then you just do order of operations

  4. freckles
    • one year ago
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    or ...

  5. freckles
    • one year ago
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    \[\text{ Let } S_{i=1 \text{ to } i=13}=\sum_{i=1}^{13}(3i+2)= \sum_{i=1}^{13}(3i) + \sum_{i=1} ^{13} 2 \\ S_{ i=1 \text{ to } i=13 }=3 \frac{13(13+1)}{2}+2(13) \] Bur we don't want to include what happens at i=1 or i=2 or i=3 or i=4 so we want to exclude the sum from i=1 to i=4 \[S_{i=5 \text{ to } i=13}=S_{i=1 \text{ to } i=13}-S_{i=1 \text{ to } i=4} \] where \[S_{i=1 \text{ to } i=4} =\sum_{i=1}^4 (3i+2)\]

  6. anonymous
    • one year ago
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    I'm confused

  7. freckles
    • one year ago
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    can you tell me on what part

  8. anonymous
    • one year ago
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    The last part that you added

  9. freckles
    • one year ago
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    like are you talking about applying the following formulas: \[\sum_{i=1}^{n} i=\frac{n(n+1)}{2} \\ \text{ and } \\ \sum_{i=1}^n c=cn\]

  10. freckles
    • one year ago
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    or are you talking about \[\sum_{i=1}^{13} f(i)=\sum_{i=1}^4 f(i)+\sum_{i=5}^{13} f(i) \\\ \]

  11. anonymous
    • one year ago
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    No, the part where you said Si=1 to I=4

  12. freckles
    • one year ago
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    I don't quite understand if you understand the last thing I asked you about then why are you asking me about the sum from i=1 to i=4

  13. freckles
    • one year ago
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    c=a+b is equivalent to the equation c-a=b or also known as b=c-a

  14. anonymous
    • one year ago
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    I'm trying to understand what you're saying ...

  15. freckles
    • one year ago
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    \[\sum_{i=1}^{13}f(i)=\sum_{i=1}^{4} f(i)+\sum_{i=5}^{13} f(i) \\ \text{ subtract } \sum_{i=1}^4 f(i) \text{ on both sides } \\ \sum_{i=1}^{13} f(i)-\sum_{i=1}^4f(i)=\sum_{i=5}^{13} f(i)\] as you see the sum from i=5 to i=13 is equal to difference of (the sum from i=1 to i=13) and (the sum from i=1 to i=4)

  16. anonymous
    • one year ago
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    I don't understand anything you're typing, but thanks anyway.

  17. freckles
    • one year ago
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    hmmm... I wonder how they expect you to evaluate the thing involving sigma if you never seen it before

  18. freckles
    • one year ago
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    \[\sum_{i=1}^{13} f(i)=f(1)+f(2)+f(3)+f(4)+ f(5)+f(6)+f(7)+f(8)+f(9)+ \\ f(10)+f(11)+f(12)+ f(13) \\ \\ \sum_{i=1}^{4} f(i)=f(1)+f(2)+f(3)+f(4) \\ \text{ as you see subtract the second line from the first }\\ \text{ (first actually occupies two lines because I couldn't fit it on one) gives } \\ f(5)+f(6)+f(7)+f(8)+f(9)+f(10)+f(11)+f(12)+f(13) \\ \text{ which is the same as } \sum_{i=5}^{13} f(i)\]

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