求积(1+1/1*3)(1+1/2*4)(1+1/3*5)......(1+1/99*101)的整数部分
答案:2 悬赏:60 手机版
解决时间 2021-03-22 15:21
- 提问者网友:像風在裏
- 2021-03-22 09:58
求积(1+1/1*3)(1+1/2*4)(1+1/3*5)......(1+1/99*101)的整数部分
最佳答案
- 五星知识达人网友:街头电车
- 2021-03-22 10:55
式子1+1/1*3=(2/1)*(2/3),
1+1/2*4=(3/2)*(3/4).
(1+1/1*3)(1+1/2*4)(1+1/3*5)......(1+1/99*101)
=[(2/1)*(2/3)]*[(3/2)*(3/4)]*[(4/3)*(4/5)]……[(100/99)*(100/101)]
=2*【(2/3)*(3/2)】*[(3/4)*(4/3)]*[4/5…………(100、99)]*(100/101)
=200/101
=1.98
1+1/2*4=(3/2)*(3/4).
(1+1/1*3)(1+1/2*4)(1+1/3*5)......(1+1/99*101)
=[(2/1)*(2/3)]*[(3/2)*(3/4)]*[(4/3)*(4/5)]……[(100/99)*(100/101)]
=2*【(2/3)*(3/2)】*[(3/4)*(4/3)]*[4/5…………(100、99)]*(100/101)
=200/101
=1.98
全部回答
- 1楼网友:走死在岁月里
- 2021-03-22 12:09
[1+1/(1*3)]*[1+1/(2*4)]*[1+1/(3*5)]*...*[1+1/(97*99)]*[1+1/(98*100)]
=(2*2/1*3)*(3*3/2*4)*(4*4/3*5)*(5*5/4*6)*...*(98*98/97*99)*(99*99/98*100)
=2/1*99/100
=1.98
=(2*2/1*3)*(3*3/2*4)*(4*4/3*5)*(5*5/4*6)*...*(98*98/97*99)*(99*99/98*100)
=2/1*99/100
=1.98
我要举报
如以上问答信息为低俗、色情、不良、暴力、侵权、涉及违法等信息,可以点下面链接进行举报!
大家都在看
推荐资讯