设{a}是正数数列,其前n项和Sn满足Sn=14(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=1sn,Tn是数列{bn}的前n项和,求limn→

题目简介

设{a}是正数数列,其前n项和Sn满足Sn=14(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=1sn,Tn是数列{bn}的前n项和,求limn→

题目详情

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
题型:解答题难度:中档来源:宣武区一模

答案

(1)由a1=S1=class="stub"1
4
(a1-1)(a1+3)
,及an>0,得a1=3
Sn=class="stub"1
4
(an-1)(an+3)
Sn-1=class="stub"1
4
(an-1-1)(an-1+3)

∴当n≥2时,an=class="stub"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=class="stub"1
Sn
=class="stub"1
2
(class="stub"1
n
-class="stub"1
n+2
)

Tn=b1+b2+…+bn
 =class="stub"1
2
(1-class="stub"1
3
+class="stub"1
2
-class="stub"1
4
++class="stub"1
n-1
-class="stub"1
n+1
+class="stub"1
n
-class="stub"1
n+2
)

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

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

an+bn
2
<0
,得a1+(b1-a1)•(class="stub"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)

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