设数列{an}的前n项积为Tn,且Tn=2-2an(n∈N*).(1)求1T1,1T2,1T3,并证明1Tn-1Tn-1=12(n≥2);(2)设bn=(1-an)(1-an+1),求数列{bn}的前

题目简介

设数列{an}的前n项积为Tn,且Tn=2-2an(n∈N*).(1)求1T1,1T2,1T3,并证明1Tn-1Tn-1=12(n≥2);(2)设bn=(1-an)(1-an+1),求数列{bn}的前

题目详情

设数列{an}的前n项积为Tn,且Tn=2-2an(n∈N*).
(1)求
1
T1
1
T2
1
T3
,并证明
1
Tn
-
1
Tn-1
=
1
2
(n≥2)

(2)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

(1)令n=1,可得T1=a1=2-2a1,可得a1=class="stub"2
3
,即T1=  class="stub"2
3

令n=2可得T2=2-2a2,即class="stub"2
3
a2=2-2a2,解得a2=class="stub"3
4
同理可求a3=class="stub"4
5

class="stub"1
T1
=class="stub"3
2
,class="stub"1
T2
=2,class="stub"1
T3
=class="stub"5
2

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
(n≥2)

(2)数列{class="stub"1
Tn
}
为等差数列,class="stub"1
Tn
=class="stub"n+2
2

当n≥2时,an=
Tn
Tn-1
=class="stub"n+1
n+2
,,当n=1时,a1=class="stub"2
3
也符合,所以an=class="stub"n+1
n+2

bn=class="stub"1
(n+2)(n+3)
=class="stub"1
n+2
-class="stub"1
n+3

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

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