(1)因为an 是等差数列
a2=3,a5=9
d=(9-3)/(5-2)=2
a1=3-2=1
所以an=1+2*(n-1)=2n-1
(2)sn+1/2bn=1/2
s(n-1)+1/2b(n-1)=1/2
两式相减,整理得：3bn=b(n-1)
bn/b(n-1)=1/3
所以bn为等比数列.
(3)因为bn=(1/3)^n
Cn=an*bn
=(2n-1)*(1/3)^n
=2n*(1/3)^n-(1/3)^n
Tn=1/2+1/2*[1-(1/3)^(n-1)]-[(6n-3)/2]*(1/3)^(n+1)