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{Snn}为等差数列;(2)设数列{1anan+1}的前n项和为Tn,证明:15≤Tn<14.-数学

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{Snn}为等差数列;(2)设数列{1anan+1}的前n项和为Tn,证明:15≤Tn<14.-数学

题目详情

设数列{an}的前n项和为Sna1=1,an=
Sn
n
+2(n-1)(n∈N+)

(1)求证:数列{
Sn
n
}
为等差数列;
(2)设数列{
1
anan+1
}
的前n项和为Tn,证明:
1
5
Tn
1
4
题型:解答题难度:中档来源:宿州模拟

答案

(1)证明:由题意:nan=Sn+2n(n-1),∴n(Sn-Sn-1)=Sn+2n(n-1)(n∈N+,n≥2)…(2分)
即:(n-1)Sn-nSn-1=2n(n-1),∴
Sn
n
-
Sn-1
n-1
=2

所以数列{
Sn
n
}
为等差数列;                                             …(6分)
(2)由(1)得:
Sn
n
=1+(n-1)×2
,∴Sn=2n2-n,
∴an=Sn-Sn-1=2n2-n-2(n-1)2+(n-1)=4n-3,(n∈N+,n≥2)…(8分)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
4
(1-class="stub"1
5
+class="stub"1
5
-class="stub"1
9
+…class="stub"1
4n-3
-class="stub"1
4n+1
)=class="stub"1
4
(1-class="stub"1
4n+1
)<class="stub"1
4
,…(10分)
又Tn为增函数,∴TnT1=class="stub"1
5
,∴class="stub"1
5
Tn<class="stub"1
4
…(13分)

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