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{
(3)在(2)的条件下,令Mn=T1•T2+T2•T3+…+Tn•Tn+1,求证:
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答案
(1)∵2Sn=1-an,∴n≥2时,2Sn-1=1-an-1,
两式相减可得an=
an-1,1 3
∵2S1=1-a1,∴a1=1 3
∴an=
;1 3n
(2)证明:∵2Tn=1-an,∴2Tn=1-
,Tn Tn-1
∴
-1 Tn
=21 Tn-1
∴{
}为等差数列;1 Tn
∵T1=a1=1 3
∴
=2n+11 Tn
∴Tn=
,an=1 2n+1
;2n-1 2n+1
(3)证明:∵Tn=
,∴TnTn+1=1 2n+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
∴
≤Mn<1 15
.1 6