已知等比数列{an}中,Sn为数列{an}的前n项和,满足Sn=k*1/3^n+1/2

(1)Sn=k/3^n+1/2,①
n=1时a1=k/3+1/2,
n>1时S=k/3^(n-1)+1/2,②
①-②,an=k/3^n-k/3^(n-1)=-2k/3^n,
∴a2=-2k/9,a2/a1=-2k/(3k+9/2)=a/an=1/3,
∴-6k=3k+9/2,k=-1/2,
an=1/3^n.
(2)anbn=(2n+1)/3^n,
Tn=1+5/3^2+7/3^3+……+(2n+1)/3^n,③
(1/3)Tn= 3/3^2+5/3^3+……+(2n-1)/3^n+(2n+1)/3^(n+1),④
③-④,(2/3)Tn=1+2(1/3^2+1/3^3+……+1/3^n)-(2n+1)/3^(n+1)
=1+2[1/9-1/3^(n+1)]/(2/3)-(2n+1)/3^(n+1)
=1+1/3-1/3^n-(2n+1)/3^(n+1)
=4/3-(2n+4)/3^(n+1),
∴Tn=2-(n+2)/3^n.