在数列{an}中,a1=1,Sn=a1+a2+........+an,an=2Sn-1(n属于N*),且n≥2).求证:(1)数列{Sn}是等比数列;(2)求数列{an}的通项公式.
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(1). an=2Sn-1 == Sn - S(n-1) = 2*S(n-1) === Sn/S(n-1) = 3== {Sn}是等比数列，公比为3。(2). Sn = S1*3^(n-1) = 3^(n-1),(S1=a1=1)== an = Sn -S(n-1) = 3^(n-1) - 3^(n-2) = 2*3^(n-2)