设{an}是公比为q的等比数列,其前n项乘积为Tn,并满足下列条件,①a1>1②a99×a100>1③(a99 -1)/(a100 -1)<0 判断下列选项对错,并说明原因⑴0<q<1⑵T198<1⑶a99×a101<1⑷使Tn<1成立的最
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/02 06:05:12
![设{an}是公比为q的等比数列,其前n项乘积为Tn,并满足下列条件,①a1>1②a99×a100>1③(a99 -1)/(a100 -1)<0 判断下列选项对错,并说明原因⑴0<q<1⑵T198<1⑶a99×a101<1⑷使Tn<1成立的最](/uploads/image/z/7124821-61-1.jpg?t=%E8%AE%BE%EF%BD%9Ban%EF%BD%9D%E6%98%AF%E5%85%AC%E6%AF%94%E4%B8%BAq%E7%9A%84%E7%AD%89%E6%AF%94%E6%95%B0%E5%88%97%2C%E5%85%B6%E5%89%8Dn%E9%A1%B9%E4%B9%98%E7%A7%AF%E4%B8%BATn%2C%E5%B9%B6%E6%BB%A1%E8%B6%B3%E4%B8%8B%E5%88%97%E6%9D%A1%E4%BB%B6%2C%E2%91%A0a1%EF%BC%9E1%E2%91%A1a99%C3%97a100%EF%BC%9E1%E2%91%A2%EF%BC%88a99+-1%EF%BC%89%EF%BC%8F%EF%BC%88a100+-1%EF%BC%89%EF%BC%9C0+%E5%88%A4%E6%96%AD%E4%B8%8B%E5%88%97%E9%80%89%E9%A1%B9%E5%AF%B9%E9%94%99%2C%E5%B9%B6%E8%AF%B4%E6%98%8E%E5%8E%9F%E5%9B%A0%E2%91%B40%EF%BC%9Cq%EF%BC%9C1%E2%91%B5T198%EF%BC%9C1%E2%91%B6a99%C3%97a101%EF%BC%9C1%E2%91%B7%E4%BD%BFTn%EF%BC%9C1%E6%88%90%E7%AB%8B%E7%9A%84%E6%9C%80)
设{an}是公比为q的等比数列,其前n项乘积为Tn,并满足下列条件,①a1>1②a99×a100>1③(a99 -1)/(a100 -1)<0 判断下列选项对错,并说明原因⑴0<q<1⑵T198<1⑶a99×a101<1⑷使Tn<1成立的最
设{an}是公比为q的等比数列,其前n项乘积为Tn,并满足下列条件,①a1>1②a99×a100>1③(a99 -1)/(a100 -1)<0 判断下列选项对错,并说明原因⑴0<q<1⑵T198<1⑶a99×a101<1⑷使Tn<1成立的最小自然数n等于199
设{an}是公比为q的等比数列,其前n项乘积为Tn,并满足下列条件,①a1>1②a99×a100>1③(a99 -1)/(a100 -1)<0 判断下列选项对错,并说明原因⑴0<q<1⑵T198<1⑶a99×a101<1⑷使Tn<1成立的最
1.0因为③(a99 -1)/(a100 -1)<0,所以③(a99 -1)(a100 -1)<0
所以a99>1,a100<1
a100=a99*q
所以q<1
又②a99×a100>1,所以a99和a100同号
所以q>0
2.不正确
T198=a1*a2*...*a198=a1*a1*q*a1*q²...a1*q^(197)
=a1^198*q(1+2+...+197)
=a1^198*q^197*99
an=a1*q^(n-1)
a99=a1*q^98
a100=a1*q^99
所以a99*a100=a1*q^98*a1*q^99=a1^2*q^197>1
所以q^197>1/a1^2
所以q^197*99>a1-198
所以T198=a1^198*q^197*99>1
3.正确
a99×a101
=a100^2<1
4.a100<1
a99*a101=(a100)^2<1
a98*a102=(a100)^2<1
...
a1*a199=(a100)^2<1
所以T199<1
n最小为199