(1)解:∵a1=
,an=-2SnSn-1,1 2
∴S1=
,S2=1 2
,S3=1 4
;1 6
(2)解:∵an=-2SnSn-1(n≥2),
∴Sn-Sn-1=-2SnSn-1,
∴
-1 Sn
=2,1 Sn-1
∴{
}是以2为首项,2为公差的等差数列.1 Sn
∴
=2n,1 Sn
∴Sn=
,1 2n
∴n≥2时,an=
,1 2n-2n2
∴an=
,n=11 2
,n≥21 2n-2n2
(3)证明:∵Sn2=
<1 4n2
(1 4
-1 n-1
)(n≥2)1 n
∴n≥2时,S12+S22+S32+…+Sn2<
+1 4
(1-1 4
+1 2
-1 2