若等比数列{an}满足anan+1=16^n
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若等比数列{an}满足anan+1=16^n
若等比数列{an}满足anan+1=16^n
若等比数列{an}满足anan+1=16^n
a(n) = a1·q^(n-1)
a(n)·a(n+1) = a1²·q^(2n-1) = a1²q·q^2n = 16^n
由于a1²·q为与常数项,所以a1·q = 1,q^2n = 16^n
解得a1 =1/2 ,q = 4
则a(n) = 4^(n-1)/2 = 4^n/8
若等比数列{an}满足anan+1=16^n
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