数列an中,an>0,且anan+1是公比为q的等比数列,满足anan+1+an+1an+2>an2an+3,则公比的取值范围

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数列an中,an>0,且anan+1是公比为q的等比数列,满足anan+1+an+1an+2>an2an+3,则公比的取值范围

数列an中,an>0,且anan+1是公比为q的等比数列,满足anan+1+an+1an+2>an2an+3,则公比的取值范围
数列an中,an>0,且anan+1是公比为q的等比数列,满足anan+1+an+1an+2>an2an+3,则公比的取值范围

数列an中,an>0,且anan+1是公比为q的等比数列,满足anan+1+an+1an+2>an2an+3,则公比的取值范围
看了半天题目.题目中的式子是不是an*a(n+1)+a(n+1)*a(n+2)>a(n+2)*a(n+3)呢?如果是这样的话,
设数列{an*a(n+1)}为{bn},那么有bn是公比为q的等比数列,且满足bn+b(n+1)>b(n+2),
即bn+bn*q>bn*q^2
∵an>0,∴bn>0
∴1+q>q^2
即q^2-q-1

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