已知等比数列{an}的公比大于1,Sn是数列{an}的前n项和,S3=39,且a1,23a2,13a3依次成等差数列.(Ⅰ)求数列{an}的通项公式;(II)若数列{bn}满足:b1=3,bn=an(

题目简介

已知等比数列{an}的公比大于1,Sn是数列{an}的前n项和,S3=39,且a1,23a2,13a3依次成等差数列.(Ⅰ)求数列{an}的通项公式;(II)若数列{bn}满足:b1=3,bn=an(

题目详情

已知等比数列{an}的公比大于1,Sn是数列{an}的前n项和,S3=39,且a1
2
3
a2
1
3
a3
依次成等差数列.
(Ⅰ)求数列{an}的通项公式;
(II)若数列{bn}满足:b1=3,bn=an
1
a1
+
1
a2
+…+
1
an-1
)(n≥2),求数列{bn}的前n项和Tn
题型:解答题难度:中档来源:杭州一模

答案

(Ⅰ)∵a1,class="stub"2
3
a2
class="stub"1
3
a3
依次成等差数列,∴class="stub"4
3
a2=a1+class="stub"1
3
a3
,即:4a2=3a1+a3.
设等比数列{an}公比为q,则4a1q=3a1+a1q2,∴q2-4q+3=0.
∴q=1(舍去),或q=3.
S3=a1+a1q+a1q2=13a1=39,故a1=3,
an=3n.              
(Ⅱ) 当n≥2时,bn=3n•(class="stub"1
3
+class="stub"1
32
+…+class="stub"1
3n-1
)=3n
class="stub"1
3
[1-(class="stub"1
3
)
n-1
]
1-class="stub"1
3
=class="stub"1
2
[3n-3]

bn=
3
class="stub"1
2
[3n-3]
n=1
n≥2

∴Tn=3+class="stub"1
2
[9+27+81+…+3n-3(n-1)]=3+class="stub"1
2
[
9(1-3n-1)
1-3
-3(n-1)]=class="stub"1
4
3n+1-class="stub"3
2
n+class="stub"9
4

Tn=class="stub"1
4
3n+1-class="stub"3
2
n+class="stub"9
4

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