问题 解答题
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
答案

(1)令n=1,可得T1=a1=2-2a1可得a1=

2
3
,即T1=  
2
3

令n=2可得T2=2-2a2,即

2
3
a2=2-2a2,解得a2=
3
4
同理可求a3=
4
5

1
T1
=
3
2
1
T2
=2,
1
T3
=
5
2

由题意可得:Tn=2-2

Tn
Tn-1
 ⇒Tn•Tn-1=2Tn-1-2Tn(n≥2),

所以

1
Tn
-
1
Tn-1
=
1
2
(n≥2);

(2)数列{

1
Tn
}为等差数列,
1
Tn
=
n+2
2

当n≥2时,an=

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

bn=

1
(n+2)(n+3)
=
1
n+2
-
1
n+3

sn

1
3×4
+
1
4×5
+…+
1
(n+2)•(n+3)
=
1
3
-
1
4
+
1
4
-
1
5
+…+
1
n+2
-
1
n+3
=
1
3
-
1
n+3
=
n
3n+9

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