已知a1=1数列{an}的前n项和Sn满足nSn+1-(n+3)Sn=0(1)求an(2)令bn=1an,求{bn}的前n项和Tn.-数学

题目简介

已知a1=1数列{an}的前n项和Sn满足nSn+1-(n+3)Sn=0(1)求an(2)令bn=1an,求{bn}的前n项和Tn.-数学

题目详情

已知a1=1数列{an}的前n项和Sn满足nSn+1-(n+3)Sn=0
(1)求an
( 2 )bn=
1
an
,求{bn}的前n项和Tn
题型:解答题难度:中档来源:不详

答案

(1)∵nSn+1-(n+3)Sn=0,即nan+1=3Sn①
∴(n-1)an=3Sn-1(n≥2)②
①-②得nan+1=(n+2)an(n≥2)
∴an=class="stub"n+1
n-1
×class="stub"n
n-2
×class="stub"n-1
n-3
×…×class="stub"6
4
×class="stub"5
3
×class="stub"4
2
×class="stub"3
1

=
n(n+1)
2
(n≥2),
a1=1也适合上式,
∴an=
n(n+1)
2
(n∈N*).
(2)bn=class="stub"1
an
=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

=class="stub"2n
n+1

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