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解答

A
bn = 2^an
b(n+1) = 2^a(n+1) = 2^(an + d) = (2^an)(2^d)
得證bn為公比r = 2^d的等比數列

B
a1 < a3
即d > 0
則r = 2^d > 1
故b3 > b2

C
d = (a3 - a1)/2
0 < a1 < 2
=> -2 < -a1 < 0
=> (4-2) < (a3 - a1) < (4-0)
=> 2/2 < (a3 - a1)/2 < 4/2
=> 1 < d < 2
=> 2 < 2^d < 4
已知b3 = 2^a3 = 16
b4 = b3*2^d
=>32 < b4 < 64

D
b2*b4 = b3^2 = 16^2 = 256

E
上面證過了

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