∵{log 2a n}是公差为-1的等差数列
∴log 2a n=log 2a 1-n+1
∴ a n = 2 ( log 2 a 1 -n+1)
∴ s n = a 1 (1+
1
2 +…+
1
2 n-1 )=
a 1 [1- (
1
2 ) 2 ]
1-
1
2 ,
lim
n→∞ S n =
5
3
∴ a 1 =
5
6 .
故选B.
∵{log 2a n}是公差为-1的等差数列
∴log 2a n=log 2a 1-n+1
∴ a n = 2 ( log 2 a 1 -n+1)
∴ s n = a 1 (1+
1
2 +…+
1
2 n-1 )=
a 1 [1- (
1
2 ) 2 ]
1-
1
2 ,
lim
n→∞ S n =
5
3
∴ a 1 =
5
6 .
故选B.