题目内容

已知数列{an}是等差数列,且a1=2,a1+a2+a3=12.

(1)求数列{an}的通项公式;

(2)令bn=an·3n,求数列{bn}的前n项和公式.

解:(1)设数列{an}的公差为d,则a1+a2+a3=3a1+3d=12.

a1=2,得d=2.

所以an=2n.

(2)由bn=an·3n=2n·3n,得

Sn=2·3+4·32+…+(2n-2)·3n-1+2n·3n,                                                                  ①

3Sn=2·32+4·33+…+(2n-2)·3n+2n·3n+1.                                                               ②

①-②得

-2Sn=2(3+32+33+…+3n)-2n·3n+1=3(3n-1)-2n·3n+1.

所以Sn=+n·3n+1.

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