1/[n(n+1)]+1/[(n+1)(n+2)]+1/[(n+2)(n+3)]+……+1/[（n+2004)(n+2005)]
=1/n-1/(n+1)+1/(n+1)-1/(n+2)+.+1/(n+2004)-1/(n+2005)
=1/n-1/(n+2005)
=2005/n(n+2005)
一道初二分式加减题
一道初二分式加减题
计算1/[n(n+1)]+1/[(n+1)(n+2)]+1/[(n+2)(n+3)]+……+1/[（n+2004)(n+2005)]
计算1/[n(n+1)]+1/[(n+1)(n+2)]+1/[(n+2)(n+3)]+……+1/[（n+2004)(n+2005)]
数学人气：816 ℃时间：2020-04-30 18:21:08
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