(1)由题意知,解得a1=3,由此能够推出数列{an}是以3为首项,2为公差的等差数列,所以an=3+2(n-1)=2n+1.
(2)由题意知Tn=3×21+5×22+…+(2n+1)•2n,2Tn=3×22+5×23+(2n-1)•2n+(2n+1)2n+1,二者相减可得到Tn=a1b1+a2b2+…+anbn的值.
【解析】
(1)当n=1时,,解出a1=3,
又4Sn=an2+2an-3①
当n≥2时4sn-1=an-12+2an-1-3②
①-②4an=an2-an-12+2(an-an-1),即an2-an-12-2(an+an-1)=0,
∴(an+an-1)(an-an-1-2)=0,
∵an+an-1>0∴an-an-1=2(n≥2),
∴数列{an}是以3为首项,2为公差的等差数列,∴an=3+2(n-1)=2n+1.
(2)Tn=3×21+5×22+…+(2n+1)•2n③
又2Tn=3×22+5×23+(2n-1)•2n+(2n+1)2n+1④
④-③Tn=-3×21-2(22+23++2n)+(2n+1)2n+1-6+8-2•2n-1+(2n+1)•2n+1=(2n-1)•2n+2