已知数列{an}的前n项和Sn满足an-1Sn=a-1a(a>0),数列{bn}满足bn=an•logaan(1)求数列{an}的通项;(2)求数列{bn}的前n项和Tn.-数学

题目简介

已知数列{an}的前n项和Sn满足an-1Sn=a-1a(a>0),数列{bn}满足bn=an•logaan(1)求数列{an}的通项;(2)求数列{bn}的前n项和Tn.-数学

题目详情

已知数列{an}的前n项和Sn满足
an-1
Sn
=
a-1
a
(a>0),数列{bn}满足bn=an•logaan
(1)求数列{an}的通项;
(2)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)当n=1时,a1=a>0且a≠1,
当n≥2时,Sn=class="stub"a
a-1
(an-1)
Sn-1=class="stub"a
a-1
(an-1-1)

两式相减得an=class="stub"a
a-1
(an-an-1)
,化为
an
an-1
=a

∴数列{an}是等比数列,an=an
(2)bn=an•lo
gana
=nan.
当a=1时,Tn=1+2+…+n=
n(1+n)
2

当a≠1时,Tn=a+2a2+3a3+…+nan,
aTn=a2+2a3+…++(n-1)an+nan+1,
∴(1-a)Tn=a+a2+…+an-nan+1=
a(an-1)
a-1
-nan+1

∴Tn=
a+(na-n-1)an+1
(a-1)2

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