问题 解答题
设{a}是正数数列,其前n项和Sn满足Sn=
1
4
(an-1)(an+3).
(1)求a1的值;求数列{an}的通项公式;
(2)对于数列{bn},令bn=
1
sn
,Tn是数列{bn}的前n项和,求
lim
n→∞
Tn
答案

(1)由a1=S1=

1
4
(a1-1)(a1+3),及an>0,得a1=3

Sn=

1
4
(an-1)(an+3)得Sn-1=
1
4
(an-1-1)(an-1+3)

∴当n≥2时,an=

1
4
(
a2n
-
a2n-1
)+2(an-an-1)

∴2(an+an-1)=(an+an-1)(an-an-1)∵an+an-1>0∴an-an-1=2,

∴{an}是以3为首项,2为公差的等差数列,∴an=2n+1

(2)由(1)知Sn=n(n+2)∴bn=

1
Sn
=
1
2
(
1
n
-
1
n+2
),

Tn=b1+b2+…+bn

 =

1
2
(1-
1
3
+
1
2
-
1
4
++
1
n-1
-
1
n+1
+
1
n
-
1
n+2
)

=

1
2
[
3
2
-
2n+3
(n+1)(n+2)
]=
3
4
-
2n+3
2(n+1)(n+2)

lim
n→∞
Tn=
lim
n→∞
[
3
4
-
2n+3
2(n+1)(n+2)
]=
3
4
(13分)

an+bn
2
<0,得a1+(b1-a1)•(
1
2
)n<0

a1+b1
2n
<-a,得
b1-a1
-a1
2n

log2

a1-b1
a1
<n

因而n满足log2

a1-b1
a1
<n的最小整数(14分)

判断题
判断题