摘要：12．设数列{an}的前n项和Sn＝2an-2n. (1)求a3.a4, (2)证明:{an+1-2an}是等比数列, (3)求{an}的通项公式． [解析] (1)因为a1＝S1,2a1＝S1+2. 所以a1＝2.S1＝2.由2an＝Sn+2n知 2an+1＝Sn+1+2n+1＝an+1+Sn+2n+1. 得an+1＝Sn+2n+1.① 所以a2＝S1+22＝2+22＝6.S2＝8. a3＝S2+23＝8+23＝16.S3＝24. a4＝S3+24＝40. (2)证明:由题设和①式知 an+1-2an＝(Sn+2n+1)-(Sn+2n)＝2n+1-2n＝2n. 所以{an+1-2an}是首项为2.公比为2的等比数列． (3)an＝(an-2an-1)+2(an-1-2an-2)+-+2n-2(a2-2a1)+2n-1a1＝(n+1)·2n-1.

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