问题 解答题
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
答案

(1)∵2Sn=1-an,∴n≥2时,2Sn-1=1-an-1

两式相减可得an=

1
3
an-1

∵2S1=1-a1,∴a1=

1
3

an=

1
3n

(2)证明:∵2Tn=1-an,∴2Tn=1-

Tn
Tn-1

1
Tn
-
1
Tn-1
=2

∴{

1
Tn
}为等差数列;

∵T1=a1=

1
3

1
Tn
=2n+1

∴Tn=

1
2n+1
an=
2n-1
2n+1

(3)证明:∵Tn=

1
2n+1
,∴TnTn+1=
1
(2n+1)(2n+3)
=
1
2
1
2n+1
-
1
2n+3

∴Mn=T1•T2+T2•T3+…+Tn•Tn+1=

1
2
[
1
3
-
1
5
+
1
5
-
1
7
+…+(
1
2n+1
-
1
2n+3
)]=
1
2
(
1
3
-
1
2n+3
)

1
15
Mn
1
6

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