=Sn2,其中Sn=
. (1)求证:对一切n∈N*,有an+12-an+1=2Sn;
(2)求数列{an}的通项公式;
(3)求证:
＜3.
=Sn2,其中Sn=. (1)求证:对一切n∈N*,有an+12-an+1=2Sn;
(2)求数列{an}的通项公式;
(3)求证:＜3.
解析:(1)由=Sn2,①
则=Sn+12.②
②-①,得an+13=Sn+12-Sn2=(Sn+1+Sn)(Sn+1-Sn)=(2Sn+an+1)an+1.
因为an+1＞0,所以an+12-an+1=2Sn.
(2)由an+12-an+1=2Sn及an2-an=2Sn-1(n≥2),
两式相减,得(an+1+an)(an+1-an)=an+1+an.
因为an+1+an＞0,所以an+1-an=1(n≥2).
当n=1,2时易得a1=1,a2=2,所以an+1-an=1(n≥1).
所以{an}是以a1=1为首项,d=1为公差的等差数列.
故an=n.
(3) =
＜1+
＜1+
=1+
=1+
=1+1+
- 
-
＜2+
＜3.