题目内容

16.已知数列{an}是等差数列,且a1=2,a1+a2+a3=12.

(Ⅰ)求数列{an}的通项公式;

(Ⅱ)令bn=an·3n,求数列{bn}前n项和的公式.

16.

(Ⅰ)解: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.

 

(Ⅱ)解:由bn=an·3n=2n3n,得

Sn=2·3+4·32+…+(2n-2)·3n1+2n·3n,                                 ①

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

将①式减去②式,得

-2Sn=2(3+32+…+3n)-2n·3n+1

=3(3n-1)-2n·3n+1

 

所以Sn=+n·3n+1.


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