设An=1/n+1＋1／n+2＋1／n﹢3＋……＋／2n﹙n属于正整数）那么An+1－An＝？
答案:5 悬赏:30
解决时间 2021-04-27 17:42
- 提问者网友：北故人
- 2021-04-27 02:03
设An=1/n+1＋1／n+2＋1／n﹢3＋……＋／2n﹙n属于正整数）那么An+1－An＝？
最佳答案
- 二级知识专家网友：飘零作归宿
- 2021-04-27 02:50
An=1/n+1＋1／n+2＋1／n﹢3＋……＋／2n
An+1=1/n+2+1/n+3+...+1/2n+1/2n+1+1/2n+2
故An+1－An＝1/2n+1+1/2n+2-1/n+1
=1/2n+1-1/2n+2
=1/[（2n+1）（2n+2）]
全部回答
- 1楼网友：懂得ㄋ、沉默
- 2021-04-27 06:50
An+1 比An多了 1/2n+1+1/2n+2 少了 1/n+1
他们之间的差 就是 1/2n+1+1/2n+2-1/n+1
- 2楼网友：短发女王川岛琦
- 2021-04-27 05:11
A(n+1)=1/(n+2)+1/(n+3)``````+1/(2n+1)+1/(2n+2)
A(n+1)-An=1/(2n+1)+1/(2n+2)-1/(n+1)
=1/[(2n+1)(2n+2)]
- 3楼网友：年轻没有失败
- 2021-04-27 05:04
∵An=1/n+1＋1／n+2＋1／n﹢3＋……＋1／2n
∴An+1=1／n+2＋1／n﹢3＋……＋1／2n+1/(2n+1)+1/2(n+1)
∴An+1－An=[1／n+2＋1／n﹢3＋……＋1／2n+1/(2n+1)+1/2(n+1)]-[1／n+2＋1／n﹢3＋……＋1／2n+1/(2n+1)+1/2(n+1)]=1/(2n+1)+1/2(n+1)-1/(n+1)
- 4楼网友：有钳、任性
- 2021-04-27 03:55
A(n+1)＝1/(n+2)＋1/(n+3)＋……＋1/(2n+2)
An＝1/(n+1)＋1/(n+2)＋……＋1/(2n)
A(n+1)－An＝1/(2n+1)＋1/(2n+2)－1/(n+1)＝1/(2n+1)－1/(2n+2)＝1/[(2n+1)(2n+2)]
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