Summer Clouds moderators
19/07/2017 at 11:52-
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 :)
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\(S=1-2+3-4+....+2009-2010\)
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 :
\(S=S'+S"=1005^2-1005.1006=1005.\left(1005-1006\right)=-1005\)
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\(S=1-2+3-4+....+2007-2008+2009-2010\)
\(=\left(1-2\right)+\left(3-4\right)+...+\left(2007-2008\right)+\left(2009-2010\right)\)
\(=-1-1-....-1-1=-\dfrac{\left(2010-1\right):1+1}{2}=-1005\)