已知数列{an}的首项a1=12,前n项和Sn=n2an.(Ⅰ)求证:an+1=nn+2an;(Ⅱ)记bn=lnSn,Tn为{bn}的前n项和,求e-Tn-n的值.-数学

题目简介

已知数列{an}的首项a1=12,前n项和Sn=n2an.(Ⅰ)求证:an+1=nn+2an;(Ⅱ)记bn=lnSn,Tn为{bn}的前n项和,求e-Tn-n的值.-数学

题目详情

已知数列{an}的首项a1=
1
2
,前n项和Sn=n2an
(Ⅰ)求证:an+1=
n
n+2
an

(Ⅱ)记bn=lnSn,Tn为{bn}的前n项和,求e-Tn-n的值.
题型:解答题难度:中档来源:不详

答案

(Ⅰ)由Sn=n2an①,得Sn+1=(n+1)2an+1②,
②-①得:an+1=(n+1)2an+1-n2an
整理得,an+1=class="stub"n
n+2
an

(Ⅱ)由an+1=class="stub"n
n+2
an
,得
an+1
an
=class="stub"n
n+2

所以an=a1×
a2
a1
×
a3
a2
×…×
an
an-1

=class="stub"1
2
×class="stub"1
3
×class="stub"2
4
×…×class="stub"n-2
n
×class="stub"n-1
n+1

=class="stub"1
n(n+1)
(n≥2),
又当n=1时,a1=class="stub"1
2
,所以an=class="stub"1
n(n+1)

Sn=n2an=class="stub"n
n+1
,bn=lnSn=lnn-ln(n+1),
∴Tn=(ln1-ln2)+(ln2-ln3)+(ln3-ln4)+…+(lnn-ln(n+1))=-ln(n+1),
e-Tn-n=eln(n+1)-n=1

更多内容推荐