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Summer Clouds moderators

19/07/2017 at 11:52
Answers
3
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Calculating value of expression: \(S=1-2+3-4+......+2009-2010\)




    List of answers
  • ...
    Searching4You 19/07/2017 at 17:29

    This problem has a lot of ways to express.

    \(S=1-2+3-4+...+2007-2008+2009-2010\)

    We mark each pair to each other until the answer of each pair is -1 :

    \(S=\left(1-2\right)+\left(3-4\right)+...+\left(2007-2008\right)+\left(2009-2010\right)\)

    From here we have the number of pairs are : \(\dfrac{\left(\dfrac{2010-1}{1}+1\right)}{2}=1005\)(pairs)

    Then the correct answer is : \(-1\cdot1005=-1005\)

    Both answers upper there are correct, I just want to donate one more way to express this math prob :)

  • ...
    Kayasari Ryuunosuke Coordinator 19/07/2017 at 12:27

    \(S=1-2+3-4+....+2009-2010\)

    Denote :

    \(S'=1+3+.....+2009\) ; \(S"=-2-4-....-2010=-\left(2+4+....+2010\right)\)

    We have :

    \(S'=\dfrac{\left[\left(2009-1\right):2+1\right].\left(2009+1\right)}{2}=\dfrac{1005.2010}{2}=1005^2\)

    \(S"=-\dfrac{\left[\left(2010-2\right):2+1\right].\left(2010+2\right)}{2}=-\dfrac{1005.2012}{2}=-1005.1006\)

    So :

    \(S=S'+S"=1005^2-1005.1006=1005.\left(1005-1006\right)=-1005\)

  • ...
    Phan Thanh Tinh Coordinator 19/07/2017 at 12:06

    \(S=1-2+3-4+....+2007-2008+2009-2010\)

    \(=\left(1-2\right)+\left(3-4\right)+...+\left(2007-2008\right)+\left(2009-2010\right)\)

    \(=-1-1-....-1-1=-\dfrac{\left(2010-1\right):1+1}{2}=-1005\)


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