设数列{an}的前n项和为Sn=2an-4,bn=log2an,cn=1/bn^2,求证：数列{an}是等比数列?

Sn=2an -4
Sn = 2 [Sn - S] - 4
Sn = 2S + 4
Sn + 4 = 2 (S + 4)
所以 Sn + 4 构成公比为 2 的等比数列
Sn + 4 = (S1 + 4)*2^(n-1)
利用 S1 = 2a1 - 4 = a1 求出 S1 = a1 = 4
Sn + 4 = (4 + 4)*2^(n-1)
Sn = 4*(2^n -1)
an = Sn - S = 4*2^n -1 - 4*2^(n-1) + 1 = 2^(n+1)
bn = log2 an = n+1
cn = 1/bn^2 = 1/(n+1)^2
因为 an = 2^(n+1),所以
a/an = 2^(n+2)/2^(n+1) = 2
所以 数列{an}是等比数列
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此题目中 bn=log2an,cn=1/bn^2 没发挥作用