Question

已知{an}为等差数列,且a1=2,a1+a2+a3=12.(1)求数列{an}的通项公式（2）令bn=an·3^n,求

Answer 1

(1)S3=3a2=12

a2=4

d=a2-a1=2

an=2n

(2)bn=2n·3^n

Sn=2·3+4·3^2+6·3^3.+2n·3^n

3Sn= 2·3^2+4·3^3+6·3^4.-2n·3^n+1

-2Sn=2·3+2·3^2+2·3^3+.2·3^n-2n·3^n+1

-Sn=3+3^2+3^3+.3^n-n·3^n+1

-Sn=[3(1-3^n)/(1-3)]-n·3^n+1

Sn=[3-3^(n+1)]/2 -3^(n+2)