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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)]

分类: 作业答案 • 2022-08-16 00:26:06
问题描述：

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)]

答

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)

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