1/x(x+1)+1/(x+1)(x+2)+1/(x+2)(x+3)+……+1/(x+99)(x+100)

来源：学生作业帮助网编辑：六六作业网时间：2024/05/10 23:59:39

1/x(x+1)+1/(x+1)(x+2)+1/(x+2)(x+3)+……+1/(x+99)(x+100)1/x(x+1)+1/(x+1)(x+2)+1/(x+2)(x+3)+……+1/(x+99)(

1/x(x+1)+1/(x+1)(x+2)+1/(x+2)(x+3)+……+1/(x+99)(x+100)
1/x(x+1)+1/(x+1)(x+2)+1/(x+2)(x+3)+……+1/(x+99)(x+100)

1/x(x+1)+1/(x+1)(x+2)+1/(x+2)(x+3)+……+1/(x+99)(x+100)
1/x(x+1)+1/(x+1)(x+2)+1/(x+2)(x+3)+……+1/(x+99)(x+100)
=[1/x-1/(x+1)]+[1/(x+1)-1/(x+2)]+[1/(x+2)-1/(x+3)]+…… +[1/(x+99)-1/(x+100)]
=1/x-1/(x+100)
=100/x(x+100)

答：
原式
=1/x-1/(x+1)+1/(x+1)-1/(x+2)+...+1/(x+99)-1/(x+100)
=1/x-1/(x+100)

=[1/x-1/(x+1)}+.....+[1/(x+99)-1/(x+100)]
=1/x-1/(x+100)
=100/x(x+100)

1/[(x+1)*(x+2)]=1/(x+1)-1/(x+2) 1/[(x+2)*(x+3)]=1/x+2-1/x+3 …1/[(x+99)*(x+100)]=1/(x+99)-1/(x+100)相消得：99/[(x+1)*(x+100

晕，居然重复。看来这题的确没有什么难度
1/x(x+1)+1/(x+1)(x+2)+1/(x+2)(x+3)+……+1/(x+99)(x+100)
=[(1/x) - 1/(x+1)]+[1/(x+1)-1/(x+2)]+……+[1/(x+99) - 1/(x+100)]
=1/x - 1/(x+100)
=(x+99)/x(x+100)

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