若等比数列{an}满足anan+1=16^n,则公比是?

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若等比数列{an}满足anan+1=16^n,则公比是?若等比数列{an}满足anan+1=16^n,则公比是?若等比数列{an}满足anan+1=16^n,则公比是?anan+1=16^n-----

若等比数列{an}满足anan+1=16^n,则公比是?
若等比数列{an}满足anan+1=16^n,则公比是?

若等比数列{an}满足anan+1=16^n,则公比是?
anan+1=16^n-----------(1)
an-1an=16^(n-1)-------(2) (n≥2)
(1)/(2)
q^2=16
q=±4
anan+1=16^n
an^2q=16^n
∴q>0
∴q=-4舍
∴q=4

4

如果是这样的话,解答如下: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 ∵

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