1）由题意得,a1=1,当n>1时,
sn=an^2/2+an/2
sn-1=a(n-1)^2/2+a(n-1)/2,
∴sn-sn-1=an^2/2-a(n-1)^2/2+an/2-a(n-1)/2
即（an+an-1)[an-a(n-1)-1]=0,由an>0知,an-a(n-1)-1=0,故an-a(n-1)=1.
所以{an}是以1为首项,1为公差的等差数列.
2)由1）得,an=n,故bn=（2an+1）/2^n=（2n+1)/2^n,
所以Tn=3/2+5/2^2+...+(2n+1)/2^n,Tn/2=3/2^2+5/2^3+...+(2n+1)/2^(n+1),两式相减得,
-Tn/2=3/2+2[1/2^2+1/2^3+...+1/2^n)-(2n+1)]/2^(n+1)
-Tn=3+2[1/2+1/2^2+...+1/2^(n-1)]-(2n+1)]/2^(n+1)=5-(2n+9)/2^(n+1)
Tn=(2n+9)/2^(n+1)-5.