Question

1/(1x3)+1/(3x5)+1/(5x7)+.......+1/2005x2007=?

例子：2/(1X3)=1-(1/3)
2/(3X5)=(1/3)-(1/5)
2/(5X7)=(1/5)-(1/7)

Answer 1

这个问题用裂项相消法求解.1/(1X3)=1/2[1-(1/3)],1/(3X5)=1/2[(1/3)-(1/5)],
1/(5X7)=1/2[(1/5)-(1/7)],.......,1/2005x2007=1/2[1/2005-1/2007].所以
1/(1x3)+1/(3x5)+1/(5x7)+.......+1/2005x2007=1/2(1-1/3+1/3-1/5+1/5-1/7+.........+1/2005-1/2007)
=1/2(1-1/2007)

Answer 2

1/(1x3)+1/(3x5)+1/(5x7)+.......+1/2005x2007
=（1-1/3+1/3-1/5+1/5-1/7+……+1/2005-1/2007）÷2
=（1-1/2007）÷2
=2006/2007÷2
=1003/2007

Answer 3

貌似楼主已有答案啊
1/2*(1-1/3+1/3-1/5+1/5-1/7.....+1/2005-1/2007)
=1/2*(1-1/2007)
=1003/2007

Answer 4

=[2/(1x3)+2/(3x5)+2/(5x7)+.......+2/2005x2007]/2
=[1-(1/3)+(1/3)-(1/5).......+1/2005-1/2007]/2
=(1-1/2007)/2
=1003/2007

Answer 5

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