题目

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. 答案：解:(1)设数列{an}的公差为d,则a1+a2+a3=3a1+3d=12.又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,                                                                  ①3Sn=2·32+4·33+…+(2n－2)·3n+2n·3n+1.                                                               ②①－②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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