a1=S1=1.
n>=2时,an=Sn-S(n-1)=n^2-(n-1)^2=2n-1,a1=1适合此式,所以an=2n-1,n为正整数.
bn=an/3^n=(2n-1)/3^n.
Tn=1/3+3/3^2+5/3^3+…+(2n-1)/3^n (1)
(1)/3得：Tn/3=1/3^2+3/3^3+5/3^4+…+(2n-1)/3^(n+1) (2)
(1)-(2)得：2Tn/3=1/3+2/3^2+2/3^3+2/3^4+…+2/3^n-(2n-1)/3^(n+1)
=-1/3+2(1/3+1/3^2+1/3^3+…+1/3^n)-(2n-1)/3^(n+1)
=-1/3+1-1/3^n-(2n-1)/3^(n+1)
=2/3-2(n+1)/3^(n+1)
所以,Tn=1-(n+1)/3^n.