问题
解答题
已知数列{an}中a1=2,an+1=2-
(Ⅰ)求证:数列{bn}是等差数列; (Ⅱ)设Sn是数列{
(Ⅲ)设Tn是数列{ (
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答案
(Ⅰ)bn+1=
=1 an+1-1
=1 1- 1 an
,而 bn=an an-1
,1 an-1
∴bn+1-bn=
-an an-1
=1.n∈N*1 an-1
∴{bn}是首项为b1=
=1,公差为1的等差数列.(5分)1 a1-1
(Ⅱ)由(Ⅰ)可知bn=n,
bn=1 3
n. ∴Sn=1 3
(1+2+…+n)=1 3
,n(n+1) 6
于是
=1 Sn
=6(6 n(n+1)
-1 n
),1 n+1
故有
+1 S1
+…+1 S2
=6(1-1 Sn
+1 2
-1 2
+…+1 3
-1 n
)1 n+1
=6(1-
)=1 n+1
.(9分)6n n+1
(Ⅲ)证明:由(Ⅰ)可知 (
)n•bn=n•(1 3
)n,1 3
则Tn=1•
+2•(1 3
)2+…+n•(1 3
)n.∴1 3
Tn=1•(1 3
)2+2•(1 3
)3+…+(n-1)(1 3
)n+n•(1 3
)n+1.1 3
则
Tn=2 3
+(1 3
)2+(1 3
)3+…+(1 3
)n-n•(1 3
)n+1=1 3
[1-(1 2
)n]-n•(1 3
)n+1,1 3
∴Tn=
-3 4
(1 4
)n-1-1 3
•(n 2
)n<1 3
. (14分)3 4