C
因为数列{ log 2( a n-1)}(n∈N *)为等差数列,∴
故设 log 2( a n+1-1)- log 2( a n-1)=d
又 a 1=3, a 2=5,故d=1
∴
,
故{ a n-1}是首项为2,公比为2的等比数列,
∴ a n-1=2 n,∴ a n=2 n+1,∴ a n+1- a n=2 n
=
则
=1
C
因为数列{ log 2( a n-1)}(n∈N *)为等差数列,∴
故设 log 2( a n+1-1)- log 2( a n-1)=d
又 a 1=3, a 2=5,故d=1
∴
,
故{ a n-1}是首项为2,公比为2的等比数列,
∴ a n-1=2 n,∴ a n=2 n+1,∴ a n+1- a n=2 n
=
则
=1