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=
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(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
=2n,22n+1-1
=an bn
,…(8分)2n+1 2n
∴Tn=
+3 2
+…+5 22
+2n-1 2n-1
,①…(9分)2n+1 2n
将上式两边同乘以
,得1 2
Tn=1 2
+3 22
+…+5 23
+2n-1 2n
②…(10分)2n+1 2n+1
①-②,得
Tn=1 2
+3 2
+2 22
+…+2 23
-2 2n
=2n+1 2n+1
-5 2
,2n+5 2n+1
即Tn=5-
.…(12分)2n+5 2n
∵n∈N*,∴
>02n+5 2n
∴Tn<5.…(13分)