正项数列{an}的前n项和Sn满足:Sn2-(n2+n-1)Sn-(n2+n)=0(1)求数列{an}的通项公式an;(2)令bn=n+1(n+2)2an2,数列{bn}的前n项和为Tn.证明:对于任

题目简介

正项数列{an}的前n项和Sn满足:Sn2-(n2+n-1)Sn-(n2+n)=0(1)求数列{an}的通项公式an;(2)令bn=n+1(n+2)2an2,数列{bn}的前n项和为Tn.证明:对于任

题目详情

正项数列{an}的前n项和Sn满足:Sn2-(n2+n-1)Sn-(n2+n)=0
(1)求数列{an}的通项公式an
(2)令b n=
n+1
(n+2)2an2
,数列{bn}的前n项和为Tn.证明:对于任意n∈N*,都有T n
5
64
题型:解答题难度:中档来源:江西

答案

(I)由Sn2-(n2+n-1)Sn-(n2+n)=0
可得,[sn-(n2+n)](sn+1)=0
∵正项数列{an},sn>0
∴sn=n2+n
于是a1=s1=2
n≥2时,an=sn-sn-1=n2+n-(n-1)2-(n-1)=2n,而n=1时也适合
∴an=2n
(II)证明:由b n=class="stub"n+1
(n+2)2an2
=class="stub"n+1
(n+2)2•4n2
=class="stub"1
16
[class="stub"1
n2
-class="stub"1
(n+2)2
]

Tn=class="stub"1
16
[1-class="stub"1
32
+class="stub"1
22
-class="stub"1
42
+…+
class="stub"1
(n-1)2
-class="stub"1
(n+1)2
+class="stub"1
n2
-class="stub"1
(n+2)2
]
=class="stub"1
16
[1+class="stub"1
4
-class="stub"1
(n+1)2
-class="stub"1
(n+2)2
]

<class="stub"1
16
(1+class="stub"1
4
)=class="stub"5
64

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