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Math Help - Sigma Notation

  1. #1
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    Sigma Notation

    Write the indicated sum in sigma notation:

    1-(1/2)+(1/3)-(1/4)+...-(1/100)

    Another one using Special Sum Formulas
    nEi=1 (2i-3)^2

    I wanted to break it up but the square is throwing me off

    btw the n is on top of the E and i=2 is below it.
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  2. #2
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    Quote Originally Posted by pakman
    Write the indicated sum in sigma notation:

    1-(1/2)+(1/3)-(1/4)+...-(1/100)

    Another one using Special Sum Formulas
    nEi=1 (2i-3)^2

    I wanted to break it up but the square is throwing me off

    btw the n is on top of the E and i=2 is below it.
    (-1)^0\frac{1}{1}+(-1)^1\frac{1}{2}+(-1)^2\frac{1}{3}+...
    Thus,
    \sum_{k=1}^{100}(-1)^{k-1}\frac{1}{k}
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  3. #3
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    Thanks but it still makes no sense to me
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  4. #4
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    Quote Originally Posted by pakman
    Thanks but it still makes no sense to me
    What is your problem?
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  5. #5
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    Quote Originally Posted by ThePerfectHacker
    What is your problem?
    I don't even know where to begin with that problem. The only problems that we've gone over in class were ones that went in sequence... this one seems out of whack I guess.
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  6. #6
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    Quote Originally Posted by pakman
    I don't even know where to begin with that problem. The only problems that we've gone over in class were ones that went in sequence... this one seems out of whack I guess.
    Well, do you even know how a sigma works???
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  7. #7
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    Quote Originally Posted by ThePerfectHacker
    Well, do you even know how a sigma works???
    I think I understand the basic concept of it, however I don't seem to have a full grasp of it once it gets a little complex. I know what each variable does... however something like what you did for my problem seems confusing. I don't see how your solution solves my problem
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  8. #8
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    Quote Originally Posted by pakman
    I think I understand the basic concept of it, however I don't seem to have a full grasp of it once it gets a little complex. I know what each variable does... however something like what you did for my problem seems confusing. I don't see how your solution solves my problem
    You start from one and you have,
    (-1)^{1-1}\frac{1}{1}=(-1)^0\frac{1}{1}=1/1---> first term good
    (-1)^{2-1}\frac{1}{2}=(-1)^1\frac{1}{2}=-1/2---> second term good
    (-1)^{3-1}\frac{1}{3}=(-1)^2\frac{1}{3}=1/3---> thrid term good.
    And so on
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  9. #9
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    Quote Originally Posted by ThePerfectHacker
    You start from one and you have,
    (-1)^{1-1}\frac{1}{1}=(-1)^0\frac{1}{1}=1/1---> first term good
    (-1)^{2-1}\frac{1}{2}=(-1)^1\frac{1}{2}=-1/2---> second term good
    (-1)^{3-1}\frac{1}{3}=(-1)^2\frac{1}{3}=1/3---> thrid term good.
    And so on
    Thanks, that helped a lot!
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  10. #10
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    You wouldn't by chance know how to solve the second problem using a Special Sum Formula?
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  11. #11
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    Nevermind I figured it out. Square it out to 4i^2-12i+9 and just break it up from there
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  12. #12
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    Hello, pakman!

    Another one using Special Sum Formulas:
    . . \sum^n_{i=1} (2i-3)^2

    I'm only guessing, but are these the Special Sum Formulas?

    . . \sum^n_{i=1}1\;=\;n \qquad \sum^n_{i=1}i\;=\;\frac{n(n+1)}{2} \qquad \sum^n_{i=1}i^2\;=\;\frac{i(i+1)(2i+1)}{6}



    We have: . \sum^n_{i=1}(2i-3)^2\;=\;\sum^n_{i=1}\left(4i^2 - 12i + 9\right)

    . . Then: . . . . 4\sum^n_{i=1}i^2 \quad\;\;- \quad\;\;12\sum^n_{i=1}i \quad+ \quad9\sum^n_{i=1}1

    . . . . . . . . . . . . \downarrow . . . . . . . . . . \downarrow . . . . . . . \downarrow

    . . = \;4\overbrace{\left[\frac{n(n+1)(2n+1)}{6}\right]} - 12\overbrace{\left[\frac{n(n+1)}{2}\right]} \;+ \;9\overbrace{[n]}

    which simplifies to: . \frac{n}{3}\left(4n^2 - 12n + 11)
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