设数列{an}的前n项和为Sn=2an-2n,
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设数列{an}的前n项和为Sn=2an-2n,
(Ⅰ)求a1,a4
(Ⅱ)证明:{an+1-2an}是等比数列;
(Ⅲ)求{an}的通项公式.
(Ⅰ)求a1,a4
(Ⅱ)证明:{an+1-2an}是等比数列;
(Ⅲ)求{an}的通项公式.
(Ⅰ)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+1①,
则a2=S1+22=2+22=6,S2=8;a3=S2+23=8+23=16,S2=24,a4=S3+24=40;
(Ⅱ)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.
(Ⅲ)an=(an-2an-1)+2(an-1-2an-2)+…+2n-2(a2-2a1)+2n-1a1=(n+1)•2n-1
由2an=Sn+2n知:2an+1=Sn+1+2n+1=an+1+Sn+2n+1,
得an+1=sn+2n+1①,
则a2=S1+22=2+22=6,S2=8;a3=S2+23=8+23=16,S2=24,a4=S3+24=40;
(Ⅱ)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.
(Ⅲ)an=(an-2an-1)+2(an-1-2an-2)+…+2n-2(a2-2a1)+2n-1a1=(n+1)•2n-1
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