设数列{bn}的n项和为Sn,且bn=1-2Sn;数列{an}为等差数列,且a5=14,a7=20,.(1)求数列{bn}的通项公式;(2)若cn=an•bn,n=1,2,3,…,Tn为数列{cn}的

题目简介

设数列{bn}的n项和为Sn,且bn=1-2Sn;数列{an}为等差数列,且a5=14,a7=20,.(1)求数列{bn}的通项公式;(2)若cn=an•bn,n=1,2,3,…,Tn为数列{cn}的

题目详情

设数列{bn}的n项和为Sn,且bn=1-2Sn;数列{an}为等差数列,且a5=14,a7=20,.
(1)求数列{bn}的通项公式;
(2)若cn=an•bn,n=1,2,3,…,Tn为数列{cn}的前n项和.求证:Tn
7
4
题型:解答题难度:中档来源:不详

答案

(1)由bn=1-2Sn,令n=1,则b1=1-2S1,又S1=b1
所以b1=class="stub"1
3
…(2分)
当n≥2时,由bn=2-2Sn,可得bn-bn-1=-2(Sn-Sn-1)=-2bn
bn
bn-1
=class="stub"1
3
…(4分)
所以{bn}是以b1=class="stub"1
3
为首项,class="stub"1
3
为公比的等比数列,
于是bn=class="stub"1
3n
…(6分)
(2)数列{an}为等差数列,公差d=class="stub"1
2
(a7-a5)=3,可得an=3n-1…(7分)
从而cn=an•bn=(3n-1)•class="stub"1
3n

∴Tn=2•class="stub"1
3
+5•class="stub"1
32
+8•class="stub"1
33
+…+(3n-1)•class="stub"1
3n

class="stub"1
3
Tn=2•class="stub"1
32
+5•class="stub"1
33
+…+(3n-4)•class="stub"1
3n
+(3n-1)•class="stub"1
3n+1

class="stub"2
3
Tn=2•class="stub"1
3
+3•+3•class="stub"1
32
+…+3•class="stub"1
3n
-class="stub"1
3
-(3n-1)•class="stub"1
3n+1
=class="stub"7
6
-class="stub"6n+7
2•3n+1
…(11分)
∴Tn=class="stub"7
4
-class="stub"6n+7
4•3n
class="stub"7
4
.…(12分)

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