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[Post New]24 Apr 2008 15:19:42 IST
    Subject: Re:Summation
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sboosy (3063)

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Olaaa!! Perrrfect answer. 539  [723 rates]
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1^2-2^2+3^2-4^2+....-2002^2+2003^2 \\ \\ \mbox{Considering in pairs}\\ \\ \ (1-2)(1+2)+(3-4)(3+4)+....(2001-2002)(2001+2002)+2003^2 \\ \\ \mbox{Taking} \ -1 \ \mbox{common in each bracket,we get} \\ \\ -1[1+2+3+....2002]+2003^2 \\ \\ = -1\left(\frac{(2002)(2003)}{2}\right)+2003^2 \\ \\ = 2003(2003-1001) = (2003)(1002) = 2007006
 this reply: 15 points  (with 3 Olaaa!! Perrrfect answer.   in 3 votes )   [?]
 
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