设Tn为数列{an}的前n项乘积,满足Tn=1-an(n∈N*)(1)设bn=1Tn,求证:数列{bn}是等差数列;(2)设cn=2n•bn,求证数列{cn}的前n项和Sn;(3)设An=Te1+Te

题目简介

设Tn为数列{an}的前n项乘积,满足Tn=1-an(n∈N*)(1)设bn=1Tn,求证:数列{bn}是等差数列;(2)设cn=2n•bn,求证数列{cn}的前n项和Sn;(3)设An=Te1+Te

题目详情

设Tn为数列{an}的前n项乘积,满足Tn=1-an(n∈N*)
(1)设bn=
1
Tn
,求证:数列{bn}是等差数列;
(2)设cn=2n•bn,求证数列{cn}的前n项和Sn
(3)设An=
Te1
+
Te2
+…
Ten
,求证:an+1-
1
2
An≤-
1
4
题型:解答题难度:中档来源:不详

答案

(1)∵Tn=1-an,an=
Tn
Tn-1
,n≥2,
Tn=1-
Tn
Tn-1
,从而class="stub"1
Tn
-class="stub"1
Tn-1
=1,(n≥2)
∴bn-bn-1=1,(n≥2)
∵T1=a1=1-a1,
a1=class="stub"1
2
b1=class="stub"1
T1
=class="stub"1
a1
=2

∴{bn}是以2为首项,1为公差的等差数列.
(2)由(1)知bn=2+(n-1)=n+1,从而cn=(n+1)•2n,
∴Sn=2•2+3•22+…+(n+1)•2n,
2Sn=2•22+3•23+…+n•2n+(n+1)•2n+1,
两式相减,得-Sn=4+(22+23+…+2n)-(n+1)•2n+1
=4+
4(1-2n-1)
1-2
-(n+1)•2n+1
=-n•2n+1,
∴Sn=n•2n+1.
(3)∵Tn=class="stub"1
bn
=class="stub"1
n+1

∴n≥2时,an=
Tn
Tn-1
=class="stub"n
n+1

a1=class="stub"1
2
,∴an=class="stub"n
n+1
,n∈N* 

An=T12+T22+…+Tn2
=class="stub"1
22
+class="stub"1
32
+…+class="stub"1
(n+1)2

class="stub"1
2×3
+class="stub"1
3×4
+…+class="stub"1
(n+1)(n+2)

=class="stub"1
2
-class="stub"1
3
+class="stub"1
3
-class="stub"1
4
+…+class="stub"1
n+1
-class="stub"1
n+2

=class="stub"1
2
-class="stub"1
n+2

=an+1-class="stub"1
2

Anan+1-class="stub"1
2

又∵当n≥2时,An=T12+T22+…+Tn2
=class="stub"1
22
+class="stub"1
32
+…+class="stub"1
(n+1)2

=class="stub"1
22
+class="stub"1
32
+…+class="stub"1
(n+1)2
<class="stub"1
22
+class="stub"1
2×3
+class="stub"1
3×4
+…+class="stub"1
n(n+1)

=class="stub"1
22
+class="stub"1
2
-class="stub"1
3
+class="stub"1
3
-class="stub"1
4
+…+class="stub"1
n
-class="stub"1
n+1

=class="stub"1
4
+class="stub"1
2
-class="stub"1
n+1
=an-class="stub"1
4

an+1-class="stub"1
2
An≤-class="stub"1
4

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