设数列{an}的前n项和为Sn,a1=1,an=Snn+2(n-1)(n∈N*).(1)求证:数列{an}为等差数列,并分别写出an和Sn关于n的表达式;(2)设数列{1anan+1}的前n项和为Tn

题目简介

设数列{an}的前n项和为Sn,a1=1,an=Snn+2(n-1)(n∈N*).(1)求证:数列{an}为等差数列,并分别写出an和Sn关于n的表达式;(2)设数列{1anan+1}的前n项和为Tn

题目详情

设数列{an}的前n项和为Sn,a1=1,an=
Sn
n
+2(n-1)(n∈N*).
(1)求证:数列{an}为等差数列,并分别写出an和Sn关于n的表达式;
(2)设数列{
1
anan+1
}的前n项和为Tn,证明:
1
5
≤Tn
1
4

(3)是否存在自然数n,使得S1+
S2
2
+
S3
3
+…+
Sn
n
-(n-1)2=2011?若存在,求出n的值;若不存在,请说明理由.
题型:解答题难度:中档来源:不详

答案

(1)证明:由an=
Sn
n
+2(n-1),得Sn=nan-2n(n-1)(n∈N*).
当n≥2时,an=Sn-Sn-1=nan-(n-1)an-1-4(n-1),即an-an-1=4,
∴数列{an}是以a1=1为首项,4为公差的等差数列.
于是,an=4n-3,Sn═2n2-n(n∈N*).
(2)证明:∵class="stub"1
anan+1
=class="stub"1
(4n-3)(4n+1)
=class="stub"1
4
(class="stub"1
4n-3
-class="stub"1
4n+1
)

∴Tn=class="stub"1
a1a2
+class="stub"1
a2a3
+…+class="stub"1
anan+1
=class="stub"1
4
[(1-class="stub"1
5
)+(class="stub"1
5
-class="stub"1
9
)+(class="stub"1
9
-class="stub"1
13
)+…+(class="stub"1
4n-3
-class="stub"1
4n+1
)]=class="stub"1
4
(1-class="stub"1
4n+1
)<class="stub"1
4

又易知Tn单调递增,
故Tn≥T1=class="stub"1
a1a2
=class="stub"1
5

所以class="stub"1
5
≤Tn<class="stub"1
4

(3)由Sn=nan-2n(n-1),得
Sn
n
=an-2(n-1)=2n-1(n∈N*),
∴S1+
S2
2
+
S3
3
+…+
Sn
n
-(n-1)2=1+3+5+7+…+(2n-1)-(n-1)2
=n2-(n-1)2=2n-1.
令2n-1=2011,得n═1006,
即存在满足条件的自然数n=1006.

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