已知数列{an}的前n和为Sn,且Sn+12an=1(1)求a1;(2)求数列{an}的通项公式;(3)设bn=12(2n-1)an,求数列{bn}的前n项和Tn.-数学

题目简介

已知数列{an}的前n和为Sn,且Sn+12an=1(1)求a1;(2)求数列{an}的通项公式;(3)设bn=12(2n-1)an,求数列{bn}的前n项和Tn.-数学

题目详情

已知数列{an}的前n和为Sn,且Sn+
1
2
an=1

(1)求a1
(2)求数列{an}的通项公式;
(3)设bn=
1
2
(2n-1)an
,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)∵sn+class="stub"1
2
an=1
,∴s1+class="stub"1
2
a1=1
,∴a1=class="stub"2
3

(2)当n≥2时,sn=-class="stub"1
2
an+1
sn-1=-class="stub"1
2
an-1+1

an=sn-sn-1=-class="stub"1
2
an+1+class="stub"1
2
an-1-1

an=class="stub"1
3
an-1

又∵a1=class="stub"2
3
≠0

an
an-1
=class="stub"1
3

an=class="stub"2
3
(class="stub"1
3
)
n-1
=class="stub"2
3n

an=class="stub"2
3n
,n∈N*

(3)∵bn=class="stub"1
2
(2n-1)an
an=class="stub"2
3n
,n∈N*

bn=class="stub"2n-1
3n
,n∈N*

Tn=1×class="stub"1
31
+3×class="stub"1
32
+5×class="stub"1
33
+…+(2n-3)×class="stub"1
3n-1
+(2n-1)×class="stub"1
3n

class="stub"1
3
T
n
=1×class="stub"1
32
+3×class="stub"1
33
+5×class="stub"1
34
+…+(2n-3)×class="stub"1
3n
+(2n-1)×class="stub"1
3n+1

class="stub"2
3
Tn=class="stub"1
31
+2(class="stub"1
32
+class="stub"1
33
+…+class="stub"1
3n
)-(2n-1)•class="stub"1
3n+1
=class="stub"2
3
-class="stub"2n+2
3n+1

Tn=1-class="stub"n+1
3n
,n∈N*

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