计算:1/(x+1)(x+2)+1/(x+2)(x+3)+1/(x+3)(x+4)+...+1/(x+2009)(x+2

来源：学生作业帮编辑：搜搜考试网作业帮分类：数学作业时间：2024/06/10 03:19:18

计算:1/(x+1)(x+2)+1/(x+2)(x+3)+1/(x+3)(x+4)+...+1/(x+2009)(x+2010)=1/2x+4020

应该是1/(x+1)(x+2)+1/(x+2)(x+3)+1/(x+3)(x+4)+...+1/(x+2009)(x+2010)=1/（2x+4020 ）吧
1/(x+1)-1/(x+2)+1/(x+2）-1/(x+3)+1/(x+3)-1/(x+4)+...+1/(x+2009)-1/(x+2010)=1/（2x+4020）
1/(x+1)-1/(x+2010)=1/[2（x+2010）]
2009/(x+1)(x+2010)=1/[2(x+2010)]
2009/(x+1)=1/2
x=4017

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