已知各项均为正数的等差数列{an}的前以项和为Sn,若S3=18,且a1+1,a2,a3成等比数列.(1)求{an}的通项公式;(2)记bn=an3n+1(n∈N*),求数列{bn}的前n项和Tn.-

题目简介

已知各项均为正数的等差数列{an}的前以项和为Sn,若S3=18,且a1+1,a2,a3成等比数列.(1)求{an}的通项公式;(2)记bn=an3n+1(n∈N*),求数列{bn}的前n项和Tn.-

题目详情

已知各项均为正数的等差数列{an}的前以项和为Sn,若S3=18,且a1+1,a2,a3成等比数列.
(1)求{an}的通项公式;
(2)记bn=
an
3n+1
(n∈N*),求数列{bn}的前n项和Tn
题型:解答题难度:中档来源:不详

答案

(1)∵S3=18,a1+1,a2,a3成等比数列
a22=(a1+1)a3
3a1+3d=18
(a1+d)2=(1+a1)(a1+2d)

解可得,d=3或d=-2(舍去),a1=3
∴an=3+3(n-1)=3n
(2)∵bn=
an
3n+1
=class="stub"3n
3n+1
=class="stub"n
3n

∴Tn=1•class="stub"1
3
+2•class="stub"1
32
+…+n•class="stub"1
3n

class="stub"1
3
Tn
=1•class="stub"1
32
+2•class="stub"1
33
+…+class="stub"n-1
3n
+class="stub"n
3n+1

两式相减可得,class="stub"2
3
Tn=class="stub"1
3
+class="stub"1
32
+…+class="stub"1
3n
-class="stub"n
3n+1
class="stub"1
3
(1-class="stub"1
3n
)
1-class="stub"1
3
-class="stub"n
3n+1

Tn=
3-3n-1
4
-class="stub"n
2•3n

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