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计算:1/a(a+1)+1/(a+1)(a+2)+1/(a+2)(a+3)+.+1/(a+2003)(a+2004)
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人气:456 ℃ 时间:2020-06-28 08:28:21
解答
1/a(a+1) + 1/(a+1)(a+2) + 1/(a+2)(a+3).1/(a+2003)(a+2004) =1/a-1/(a+1)+1/(a+1)-1/(a+2)+1/(a+2)-1/(a+3).+1/(a+2003)-1/(a+2004) =1/a-1/(a+2004) =2004/a(a+2004) ...
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