在数列{an}中,a1=1,当n≥2时,其前n项和Sn满足Sn(Sn-an)+2an=0(Ⅰ)证明数列{1Sn}是等差数列;(Ⅱ)求Sn和数列{an}的通项公式an;(Ⅲ)设bn=Snn,求数列{bn

题目简介

在数列{an}中,a1=1,当n≥2时,其前n项和Sn满足Sn(Sn-an)+2an=0(Ⅰ)证明数列{1Sn}是等差数列;(Ⅱ)求Sn和数列{an}的通项公式an;(Ⅲ)设bn=Snn,求数列{bn

题目详情

在数列{an}中,a1=1,当n≥2时,其前n项和Sn满足Sn(Sn-an)+2an=0
(Ⅰ)证明数列{
1
Sn
}是等差数列;
(Ⅱ)求Sn和数列{an}的通项公式an
(Ⅲ)设b n=
Sn
n
,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

证明:(I)∵当n≥2时,an=Sn-Sn-1,且Sn(Sn-an)+2an=0
∴Sn[Sn-(Sn-Sn-1)]+2(Sn-Sn-1)=0
即Sn•Sn-1+2(Sn-Sn-1)=0
class="stub"1
Sn
-class="stub"1
Sn-1
=class="stub"1
2

又∵S1=a1=1,故数列{class="stub"1
Sn
}是以1为首项,以class="stub"1
2
为公差的等差数列
(II)由(I)得:class="stub"1
Sn
=class="stub"n+1
2

∴Sn=class="stub"2
n+1

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

∵n=1时,class="stub"-2
n(n+1)
无意义
故an=
1,n=1
class="stub"-2
n(n+1)
,n≥2

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

∴Tn=2(1-class="stub"1
2
+class="stub"1
2
-class="stub"1
3
+…+class="stub"1
n
-class="stub"1
n+1
)=2(1-class="stub"1
n+1
)=class="stub"2n
n+1

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