已知数列{an}的前n项和为Sn,前n项积为Tn.(1)若2Sn=1-an,n∈N+,求an.(2)若2Tn=1-an,an≠0,证明{1Tn}为等差数列,并求an.(3)在(2)的条件下,令Mn=T

题目简介

已知数列{an}的前n项和为Sn,前n项积为Tn.(1)若2Sn=1-an,n∈N+,求an.(2)若2Tn=1-an,an≠0,证明{1Tn}为等差数列,并求an.(3)在(2)的条件下,令Mn=T

题目详情

已知数列{an}的前n项和为Sn,前n项积为Tn
(1)若2Sn=1-an,n∈N+,求an
(2)若2Tn=1-an,an≠0,证明{
1
Tn
}为等差数列,并求an
(3)在(2)的条件下,令Mn=T1•T2+T2•T3+…+Tn•Tn+1,求证:
1
15
Mn
1
6
题型:解答题难度:中档来源:不详

答案

(1)∵2Sn=1-an,∴n≥2时,2Sn-1=1-an-1,
两式相减可得an=class="stub"1
3
an-1,
∵2S1=1-a1,∴a1=class="stub"1
3

an=class="stub"1
3n

(2)证明:∵2Tn=1-an,∴2Tn=1-
Tn
Tn-1

class="stub"1
Tn
-class="stub"1
Tn-1
=2
∴{class="stub"1
Tn
}为等差数列;
∵T1=a1=class="stub"1
3

class="stub"1
Tn
=2n+1
∴Tn=class="stub"1
2n+1
an=class="stub"2n-1
2n+1

(3)证明:∵Tn=class="stub"1
2n+1
,∴TnTn+1=class="stub"1
(2n+1)(2n+3)
=class="stub"1
2
class="stub"1
2n+1
-class="stub"1
2n+3

∴Mn=T1•T2+T2•T3+…+Tn•Tn+1=class="stub"1
2
[class="stub"1
3
-class="stub"1
5
+class="stub"1
5
-class="stub"1
7
+…+(class="stub"1
2n+1
-class="stub"1
2n+3
)]=class="stub"1
2
(class="stub"1
3
-class="stub"1
2n+3
)

class="stub"1
15
Mn<class="stub"1
6

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