问题
选择题
设等比数列{an}为1,2,4,8,…,其前n项和为Sn,则
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答案
∵Sn =
=2n-1,an=2n-1,1×(1-2n) 1-2
∴lim n→∞
=an Sn lim n→∞
=2n-1 2n-1 lim n→∞
=
-1 2 1 2n 1- 1 2n
.1 2
故选B.
设等比数列{an}为1,2,4,8,…,其前n项和为Sn,则
|
∵Sn =
=2n-1,an=2n-1,1×(1-2n) 1-2
∴lim n→∞
=an Sn lim n→∞
=2n-1 2n-1 lim n→∞
=
-1 2 1 2n 1- 1 2n
.1 2
故选B.