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Math Help - sum problem

  1. #1
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    sum problem

    I posted before several weeks this problem
    Determine the following sum:
    1-4+9-16+25-...-10000+10201

    i got to solutions from the forum members
    1 - S \;=\;\sum^{101}_{k=1}k^2 - 2\sum^{50}_{k=1}(2k)^2 \;=\;\sum^{101}_{k=1}k^2 - 8\sum^{50}_{k=1}k^2
    2-  \sum_{n=1}^{51}\{(2n-1)^2\} - \sum_{i=1}^{50} \{(2i)^2\}

    but they give different answers
    i want to know how to get the sigma notation
    is it by guessing or what?
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  2. #2
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    Quote Originally Posted by Abbas View Post
    I posted before several weeks this problem
    Determine the following sum:
    1-4+9-16+25-...-10000+10201

    i got to solutions from the forum members
    1 - S \;=\;\sum^{101}_{k=1}k^2 - 2\sum^{50}_{k=1}(2k)^2 \;=\;\sum^{101}_{k=1}k^2 - 8\sum^{50}_{k=1}k^2
    2-  \sum_{n=1}^{51}\{(2n-1)^2\} - \sum_{i=1}^{50} \{(2i)^2\}

    but they give different answers
    i want to know how to get the sigma notation
    is it by guessing or what?
    lets take this:

    1-4+9-16+25-...-10000+10201

    and seperate it into:

    (1+9+25+....+10201) - (4+16+......+10000)

    which is just:

    (sum of squares of odd numbers) - (sum of squares of even numbers)

    which leads to the second solution...& we should sum it up to 101 & 100 respectively as the last terms of the first & second parts are (101)^2 & (100)^2 sp (2n-1)=101 & (2n)=100

    (and thats exactly how you get the sum as you get n=51 for the first & n=50 for the second)

     \sum_{n=1}^{51}\{(2n-1)^2\} - \sum_{n=1}^{50} \{(2n)^2\}
    Last edited by adhyeta; May 3rd 2009 at 11:22 AM.
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  3. #3
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    is it right mathematically to seprate them like this?
    u'll expand the first part ( complete square ) and the find the sum and subtract with the other sum?
    thanx alot
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  4. #4
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    Jun 2008
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    it is certainly. as it will lead us back to:

    (1+9+25+....+10201) - (4+16+......+10000)
    (as we are subtracting the sums)

    which again is:

    (1-4+6-16+.................)

    which is your problem. hope you got it.

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