已知各项均为正实数的数列{an}的前n项和为Sn,4Sn=an2+2an-3对于一切n∈N*成立.(Ⅰ)求a1;(Ⅱ)求数列{an}的通项公式;(Ⅲ)设bn=2an-1,Tn为数列{anbn}的前n项

题目简介

已知各项均为正实数的数列{an}的前n项和为Sn,4Sn=an2+2an-3对于一切n∈N*成立.(Ⅰ)求a1;(Ⅱ)求数列{an}的通项公式;(Ⅲ)设bn=2an-1,Tn为数列{anbn}的前n项

题目详情

已知各项均为正实数的数列{an}的前n项和为Sn,4Sn=an2+2an-3对于一切n∈N*成立.
(Ⅰ)求a1
(Ⅱ)求数列{an}的通项公式;
(Ⅲ)设bn=
2an-1
Tn
为数列{
an
bn
}
的前n项和,求证Tn<5.
题型:解答题难度:中档来源:武昌区模拟

答案

(Ⅰ)当n=1时,4S1=4a1=a1 2+2a1-3,,得a12-4a1-3=0,
a1=3或a1=-1,由条件an>0,所以a1=3.      …(2分)
(Ⅱ)当n≥2时,4Sn=an2+2an-3,4sn-1=an-12+2an-1-3;
则4Sn-4Sn-1=an2+2an-3-an-12-2an-1+3,
所以4an=an2+2an-an-12-2an-1,an2-2an-an-12-2an-1=0,
(an+an-1)(an-an-1-2)=0,…(4分)
由条件an+an-1>0,所以an-an-1=2,…(5分)
故正数列{an}是首项为3,公差为2的等差数列,
所以an=2n+1.   …(6分)
(Ⅲ)由(Ⅰ)bn=
2an-1
=
22n+1-1
=2n,
an
bn
=class="stub"2n+1
2n
,…(8分)
∴Tn=class="stub"3
2
+class="stub"5
22
+…+class="stub"2n-1
2n-1
+class="stub"2n+1
2n
,①…(9分)
将上式两边同乘以class="stub"1
2
,得
class="stub"1
2
Tn=class="stub"3
22
+class="stub"5
23
+…+class="stub"2n-1
2n
+class="stub"2n+1
2n+1
        ②…(10分)
①-②,得
class="stub"1
2
Tn=class="stub"3
2
+class="stub"2
22
+class="stub"2
23
+…+class="stub"2
2n
-class="stub"2n+1
2n+1
=class="stub"5
2
-class="stub"2n+5
2n+1

即Tn=5-class="stub"2n+5
2n
.…(12分)
∵n∈N*,∴class="stub"2n+5
2n
>0
∴Tn<5.…(13分)

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