数列{an}的前n项和为Sn,且Sn=32(an-l),数列{bn}满足bn=14bn-1-34(n≥2),b1=3.(1)求数列{an}与{bn}的通项公式.(2)设数列{cn}满足cn=anlog

题目简介

数列{an}的前n项和为Sn,且Sn=32(an-l),数列{bn}满足bn=14bn-1-34(n≥2),b1=3.(1)求数列{an}与{bn}的通项公式.(2)设数列{cn}满足cn=anlog

题目详情

数列{an}的前n项和为Sn,且Sn=
3
2
(an-l),数列{bn}满足bn=
1
4
bn-1-
3
4
(n≥2),b1=3.
(1)求数列{an}与{bn}的通项公式.
(2)设数列{cn} 满足cn=anlog2(bn+1),其前n项和为Tn,求Tn
题型:解答题难度:中档来源:不详

答案

(1)对于数列{an},当n=1时,a1=S1=class="stub"3
2
(a1-1)
,解得a1=3.
当n≥2时,an=Sn-Sn-1=class="stub"3
2
(an-1)-class="stub"3
2
(an-1-1)
,化为an=3an-1.
∴数列{an}是首项为3,公比为3的等比数列,
an=3×3n-1=3n
对于数列{bn}满足bn=class="stub"1
4
bn-1-class="stub"3
4
(n≥2),b1=3.
可得bn+1=class="stub"1
4
(bn-1+1)

∴数列{bn+1}是以b1+1=4为首项,class="stub"1
4
为公比的等比数列.
bn+1=4×(class="stub"1
4
)n-1
,化为bn=42-n-1
(2)cn=3n•lo
g(42-n-1+1)2
=3n(4-2n)
Tn=2×31+0+(-2)•33+…+(4-2n)•3n.
3Tn=2×32+0+(-2)×34+…+(6-2n)•3n+(4-2n)•3n+1.
-2Tn=6+(-2)•32+(-2)•33+…+(-2)•3n-(4-2n)•3n+1
=6-2×
32(3n-1-1)
3-1
-(4-2n)•3n+1.
Tn=-class="stub"15
2
+(class="stub"5
2
-n)•3n+1

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