已知等差数列{a n }的首项为a,公差为d,且方程ax 2 -3x+2=0的解为1,d.

(1) a _n =2n-1 S _n =n ² (2) T _n =1+(n-1)·3 ⁿ


解:(1)方程ax ² -3x+2=0的两根为1,d.
所以a=1,d=2.
由此知a _n =1+2(n-1)=2n-1,前n项和S _n =n ² .
(2)令b _n =3 ^n-1 a _n =(2n-1)·3 ^n-1 ,
则T _n =b ₁ +b ₂ +b ₃ +…+b _n =1·1+3·3+5·3 ² +…+(2n-1)·3 ^n-1 ,
3T _n =1·3+3·3 ² +5·3 ² +…+(2n-3)·3 ^n-1 +(2n-1)·3 ⁿ ,
两式相减,得-2T _n =1+2·3+2·3 ² +…+2·3 ^n-1 -(2n-1)·3 ⁿ =1+ -(2n-1)·3 ⁿ =-2-2(n-1)·3 ⁿ .
∴T _n =1+(n-1)·3 ⁿ .