1-2+3-4+5.+2011 1-2+3-4+6.-2010

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1-2+3-4+5.+2011 1-2+3-4+6.-2010
1-2+3-4+5.+2011 1-2+3-4+6.-2010

1-2+3-4+5.+2011 1-2+3-4+6.-2010
1-2+3-4+5.+2011
=(1-2)+(3-4)+(5-6).(2009-2010)+2011
=-1-1-1...-1+2011
=-1005+2011
=1006
1-2+3-4+5-6.-2010
=(1-2)+(3-4)+(5-6).(2009-2010)
=-1-1-1...-1
=-1005

-1005

1-2+3-4+5..........+2011
=(1-2)+(3-4)+...(1999-2000)+2011
=-1-1-...-1+2011
=-1*2000/2+2011
=-1000+2011
=1011
1-2+3-4+5..............-2010
==-1-1-...-1
=-1*2000/2
=-1000

先加后减 即 先求1+3+5...... + 2011 = ((1+2011)*((2011-1)/2+1))/2=
再求减法的 -(2+4+6+....+2010) 一样算 就是等差数列求和啊。。。

1-2+3-4+5..........+2011
=(1-2)+(3-4)+...(1999-2000)+2011
=-1-1-...-1+2011
=-1*2000/2+2011
=-1000+2011
=1011
1-2+3-4+5..............-2010
==-1-1-...-1
=-1*2000/2
=-1000 我觉得应该是这样!