问题
解答题
已知等比数列{an}中,a1=
(Ⅰ)Sn为数列{an}的前n项和,求Sn; (Ⅱ)设bn=log2a1+log2a2+…+log2an,求数列{bn}的通项公式. |
答案
(Ⅰ)∵等比数列{an}的首项a1=
,公比q=1 2
.1 2
∴Sn=
=a1(1-qn) 1-q
=1-
(1-1 2
)1 2n 1- 1 2 1 2n
(Ⅱ)bn=log2a1+log2a2+…+log2an
=log2
+log21 2
+…+log21 22 1 2n
=-(1+2+3…n)
=-n(n+1) 2
所以数列{bn}的通项公式bn=-n(n+1) 2