设Sn是数列[an}的前n项和,a1=1,S2n=an(Sn-12),(n≥2).(1)求{an}的通项;(2)设bn=Sn2n+1,求数列{bn}的前n项和Tn.-数学

题目简介

设Sn是数列[an}的前n项和,a1=1,S2n=an(Sn-12),(n≥2).(1)求{an}的通项;(2)设bn=Sn2n+1,求数列{bn}的前n项和Tn.-数学

题目详情

设Sn是数列[an}的前n项和,a1=1,
S2n
=an(Sn-
1
2
),(n≥2)

(1)求{an}的通项;
(2)设bn=
Sn
2n+1
,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)∵
S2n
=an(Sn-class="stub"1
2
)

∴n≥2时,
S2n
=(Sn-Sn-1)(Sn-class="stub"1
2
)

展开化简整理得,Sn-1-Sn =2Sn-1Sn,∴class="stub"1
Sn
-class="stub"1
Sn-1
=2
,∴数列{class="stub"1
sn
 }是以2为公差的等差数列,其首项为class="stub"1
S1
=1

class="stub"1
Sn
=1+2(n-1)
Sn=class="stub"1
2n-1

由已知条件
S2n
=an(Sn-class="stub"1
2
)
可得 an=
2
S2n
2Sn-1
=
1,n=1
class="stub"-2
(2n-1)(2n-3)
,n≥2

(2)由于 bn=
Sn
2n+1
=class="stub"1
(2n-1)(2n+1)
=class="stub"1
2
(class="stub"1
2n-1
-class="stub"1
2n+1
)

∴数列{bn}的前n项和 Tn=class="stub"1
2
[(1-class="stub"1
3
)+(class="stub"1
3
-class="stub"1
5
)+(class="stub"1
5
-class="stub"1
7
)+…+(class="stub"1
2n-1
-class="stub"1
2n+1
)]

Tn=class="stub"1
2
(1-class="stub"1
2n+1
)=class="stub"n
2n+1

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