(1)∵等差数列{a n }的公差d>0,a 2 、a 5 且是方程x 2 -12x+27=0的两根, ∴a 2 =3,a 5 =9. ∴d=
∴a n =a 2 +(n-2)d=3+2(n-2)=2n-1; 又数列{b n }中,T n =1-
∴T n+1 =1-
②-①得:
∴b 1 =
∴数列{b n }是以
∴b n =
综上所述,a n =2n-1,b n =
(2)∵c n =a n ?b n =(2n-1)?
∴S n =a 1 b 1 +a 2 b 2 +…+a n b n =1×
∴
∴③-④得:
S n =1+2[
=1+2×
=2-
=2-(2n+2)× (
|