数列{an}中,an>0,且{anan+1}是公比为q(q>0)的等比数列,满足ana(n+1)+a(n+1)a(n+2)>a(n+2)a(n+3).(n∈N*),则公比q的取值范围是( )A.0<q<1+根号2/2 B.0<q<1+根号5/2C.0<q< -1+根号2/2 D.0<q< -
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![数列{an}中,an>0,且{anan+1}是公比为q(q>0)的等比数列,满足ana(n+1)+a(n+1)a(n+2)>a(n+2)a(n+3).(n∈N*),则公比q的取值范围是( )A.0<q<1+根号2/2 B.0<q<1+根号5/2C.0<q< -1+根号2/2 D.0<q< -](/uploads/image/z/13940188-52-8.jpg?t=%E6%95%B0%E5%88%97%7Ban%7D%E4%B8%AD%2Can%EF%BC%9E0%2C%E4%B8%94%7Banan%2B1%7D%E6%98%AF%E5%85%AC%E6%AF%94%E4%B8%BAq%28q%EF%BC%9E0%29%E7%9A%84%E7%AD%89%E6%AF%94%E6%95%B0%E5%88%97%2C%E6%BB%A1%E8%B6%B3ana%EF%BC%88n%2B1%EF%BC%89%2Ba%EF%BC%88n%2B1%EF%BC%89a%EF%BC%88n%2B2%EF%BC%89%EF%BC%9Ea%EF%BC%88n%2B2%EF%BC%89a%EF%BC%88n%2B3%EF%BC%89.%28n%E2%88%88N%2A%29%2C%E5%88%99%E5%85%AC%E6%AF%94q%E7%9A%84%E5%8F%96%E5%80%BC%E8%8C%83%E5%9B%B4%E6%98%AF%28+%29A.0%EF%BC%9Cq%EF%BC%9C1%2B%E6%A0%B9%E5%8F%B72%2F2+B.0%EF%BC%9Cq%EF%BC%9C1%2B%E6%A0%B9%E5%8F%B75%2F2C.0%EF%BC%9Cq%EF%BC%9C+-1%2B%E6%A0%B9%E5%8F%B72%2F2+D.0%EF%BC%9Cq%EF%BC%9C+-)
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数列{an}中,an>0,且{anan+1}是公比为q(q>0)的等比数列,满足ana(n+1)+a(n+1)a(n+2)>a(n+2)a(n+3).(n∈N*),则公比q的取值范围是( )A.0<q<1+根号2/2 B.0<q<1+根号5/2C.0<q< -1+根号2/2 D.0<q< -
数列{an}中,an>0,且{anan+1}是公比为q(q>0)的等比数列,
满足ana(n+1)+a(n+1)a(n+2)>a(n+2)a(n+3).(n∈N*),则公比q的取值范围是( )
A.0<q<1+根号2/2 B.0<q<1+根号5/2
C.0<q< -1+根号2/2 D.0<q< -1+根号5/2
数列{an}中,an>0,且{anan+1}是公比为q(q>0)的等比数列,满足ana(n+1)+a(n+1)a(n+2)>a(n+2)a(n+3).(n∈N*),则公比q的取值范围是( )A.0<q<1+根号2/2 B.0<q<1+根号5/2C.0<q< -1+根号2/2 D.0<q< -
把{anan+1}看成{bn},这样方便些.
那么根据条件就有
bn+b(n+1)>b(n+2)
即bn+q*bn>q^2 * bn
因为bn>0
所以可以约掉
得1+q>q^2
又因为q>0
所以可解得B选项.
直接用anan+1和q把式子表示出来,再两边约分得:1+q>q^2
解得q的取值范围是B
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