设{an}是正数数列,其前n项和Sn满足Sn=14(an-1)(an+3).(1)求a1的值;(3)求数列{an}的通项公式;(5)对于数列{bn},Tn为数列{bn}的前n项和,令bn=1sn,试求

题目简介

设{an}是正数数列,其前n项和Sn满足Sn=14(an-1)(an+3).(1)求a1的值;(3)求数列{an}的通项公式;(5)对于数列{bn},Tn为数列{bn}的前n项和,令bn=1sn,试求

题目详情

设{an}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;
(3)求数列{an}的通项公式;
(5)对于数列{bn},Tn为数列{bn}的前n项和,令bn=
1
sn
,试求Tn的表达式.
题型:解答题难度:中档来源:不详

答案

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

(2)由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,
∴由(1)知,{an}是以3为首项,2为公差的等差数列,∴an=2n+1.

(3)由(2)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)

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