设数列{an}的前n项积为Tn,且Tn=2-2an(n∈N*).(Ⅰ)求证数列{1Tn}是等差数列;(Ⅱ)设bn=(1-an)(1-an+1),求数列{bn}的前n项和Sn.-数学

题目简介

设数列{an}的前n项积为Tn,且Tn=2-2an(n∈N*).(Ⅰ)求证数列{1Tn}是等差数列;(Ⅱ)设bn=(1-an)(1-an+1),求数列{bn}的前n项和Sn.-数学

题目详情

设数列{an}的前n项积为Tn,且Tn=2-2an(n∈N*).
(Ⅰ)求证数列{
1
Tn
}
是等差数列;
(Ⅱ)设bn=(1-an)(1-an+1),求数列{bn}的前n项和Sn
题型:解答题难度:中档来源:不详

答案

(Ⅰ)∵Tn=2-2an
∴T1=2-2T1
T1=class="stub"2
3

class="stub"1
T1
=class="stub"3
2
(1分)
由题意可得:Tn=2-2
Tn
Tn-1
 ⇒
Tn•Tn-1=2Tn-1-2Tn(n≥2),
所以class="stub"1
Tn
-class="stub"1
Tn-1
=class="stub"1
2
(6分)
∴数列{class="stub"1
Tn
}
是以class="stub"1
2
为公差,以class="stub"3
2
为首项的等差数列
(Ⅱ)∵数列{class="stub"1
Tn
}
为等差数列,
class="stub"1
Tn
=class="stub"n+2
2

an=class="stub"n+1
n+2
,(8分)
bn=class="stub"1
(n+2)(n+3)
(10分),
Sn=class="stub"1
3×4
+class="stub"1
4×5
+…+class="stub"1
(n+2)×(n+3)
=(class="stub"1
3
-class="stub"1
4
)+(class="stub"1
4
-class="stub"1
5
)+…+(class="stub"1
n+2
-class="stub"1
n+3
)
=class="stub"1
3
-class="stub"1
n+3
=class="stub"n
3n+9
(12分)

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