数列{an}的前n项和为Sn=2n+1-2,数列{bn}是首项为a1,公差为d(d≠0)的等差数列,且b1,b3,b11成等比数列.(1)求数列{an}与{bn}的通项公式;(2)设cn=bnan,求

题目简介

数列{an}的前n项和为Sn=2n+1-2,数列{bn}是首项为a1,公差为d(d≠0)的等差数列,且b1,b3,b11成等比数列.(1)求数列{an}与{bn}的通项公式;(2)设cn=bnan,求

题目详情

数列{an}的前n项和为Sn=2n+1-2,数列{bn}是首项为a1,公差为d(d≠0)的等差数列,且b1,b3,b11成等比数列.
(1)求数列{an}与{bn}的通项公式;
(2)设cn=
bn
an
,求数列{cn}的前n项和Tn
题型:解答题难度:中档来源:佛山一模

答案

解析:(1)当n≥2时,an=Sn-Sn-1=2n+1-2n=2n,
a1=S1=21+1-2=2,也满足上式,
所以数列{an}的通项公式为an=2n
b1=a1=2,设公差为d,由b1,b3,b11成等比数列,
得(2+2d)2=2×(2+10d),化为d2-3d=0.
解得d=0(舍去)d=3,
所以数列{bn}的通项公式为bn=3n-1.
(2)由(1)可得Tn=class="stub"2
21
+class="stub"5
22
+class="stub"8
23
+…+class="stub"3n-1
2n

∴2Tn=2+class="stub"5
21
+class="stub"8
22
+…+class="stub"3n-1
2n-1

两式相减得Tn=2+class="stub"3
21
+class="stub"3
22
+…+class="stub"3
2n-1
-class="stub"3n-1
2n

=2+
class="stub"3
2
(1-class="stub"1
2n-1
)
1-class="stub"1
2
-class="stub"3n-1
2n
=5-class="stub"3n+5
2n

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