问题
解答题
设数列{an}的前n项积为Tn,且Tn=2-2an(n∈N*). (1)求
(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=
,即T1= 2 3
;2 3
令n=2可得T2=2-2a2,即
a2=2-2a2,解得a2=2 3
,同理可求a3=3 4 4 5
=1 T1
,3 2
=2,1 T2
=1 T3
;5 2
由题意可得:Tn=2-2
⇒Tn•Tn-1=2Tn-1-2Tn(n≥2),Tn Tn-1
所以
-1 Tn
=1 Tn-1
(n≥2);1 2
(2)数列{
}为等差数列,1 Tn
=1 Tn
,n+2 2
当n≥2时,an=
=Tn Tn-1
,,当n=1时,a1=n+1 n+2
也符合,所以an=2 3
.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