设公差为d
(1)
a3+a6=a2+d+a2+4d=2a2+5d=27
d=(27-2a2)/5=(27-2×6)/5=3
an=a1+(n-1)d=a2+(n-2)d=6+3(n-2)=3n
数列{an}的通项公式为an=3n
(2)
Sn=a1+a2+...+an=3(1+2+...+n)=3n(n+1)/2
Tn=Sn/[3×2^(n-1)]=n(n+1)/(3×2ⁿ)
T(n+1)/Tn=[(n+1)(n+2)/(3×2^(n+1)]/[n(n+1)/(3×2ⁿ)]
=(n+2)/(2n)
=1/2 +1/n
n=1时,T(n+1)/Tn=3/2>0
n=2时,T(n+1)/Tn=1
n≥3时,T(n+1)/Tn