题目内容
已知数列{an}的前n项和为Sn,3Sn=an-1(n∈N?).
(1)求a1,a2;
(2)求证:数列{an}是等比数列;
(3)求an和Sn.
(1)求a1,a2;
(2)求证:数列{an}是等比数列;
(3)求an和Sn.
(1)a1=-. a2=(2)见解析(3)
(1)解:由3S1=a1-1,得3a1=a1-1,∴a1=-.
又3S2=a2-1,即3a1+3a2=a2-1,得a2=.
(2)证明:当n≥2时,an=Sn-Sn-1= (an-1)- (an-1-1),得,所以{an}是首项为-,公比为-的等比数列.
(3)解:由(2)可得an=n,Sn=.
又3S2=a2-1,即3a1+3a2=a2-1,得a2=.
(2)证明:当n≥2时,an=Sn-Sn-1= (an-1)- (an-1-1),得,所以{an}是首项为-,公比为-的等比数列.
(3)解:由(2)可得an=n,Sn=.
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