1-1/1*2-1/2*3-1/3*4-1/4*5……-1/2008*2009的值如题,带过程

1/1*2=1/1-1/2
1/2*3=1/2-1/3
1/3*4=1/3-1/4
...
所以
原式=1-1/1+1/2-1/2+1/3-.....-1/2008+1/2009
=1/2009

1-1/1*2-1/2*3-1/3*4-1/4*5……-1/2008*2009
=1-（1-1/2)-(1/2-1/3)-(1/3-1/4)……-（1/2008-1/2009)
=1-1+1/2-1/2+1/3-1/3+1/4……-1/2008+1/2009
=0+1/2009
=1/2009

你把每一项都分成两项，比如
1/2*3=1/2-1/3
然后最后剩下最后一项的1/2009
最后的答案就是1/2009

1-1/1*2-1/2*3-1/3*4-1/4*5……-1/2008*2009
=1-(1/1*2+1/2*3+1/3*4+……+1/2009*2010)
=1-(1-1/2+1/2-1/3+1/3-1/4+……+1/2009-1/2010)
=1-(1-1/2010)
=1-2009/2010
=1/2010

1-1/1*2-1/2*3-1/3*4-1/4*5……-1/2008*2009
=1-(1/1*2+1/2*3+1/3*4+1/4*5……+1/2008*2009)
=1-(1-1/2)+(1/2-1/3)+(1/3-1/4)+……+(1/2008-1/2009)
=1-(1-1/2+1/2-1/3+1/3-1/4+……+1/2008-1/2009)
中间正负抵消
=1-(1-1/2009)
=1-2008/2009
=1/2009