摘要:19. 设数列{an}满足a1 = 2.an +1 = 3an-2.n = 1.2.3.-. (I)求证:数列{an-1}是等比数列, (II)求{an}的通项公式, (III)求{an}的前n项和Sn. 解:(I)证明:∵ an +1 = 3an-2.且a1 = 2. ∴ an +1-1 = 3 (an-1).且an≠1. ∴ = 3.∴数列{an-1}是等比数列. (II)∵数列{an-1}是等比数列. ∴ an-1 = (a1-1)·qn-1 = (2-1)·3n-1 = 3n-1. ∴ an = 3n-1 + 1. ∴ {an}的通项公式an = 3n-1 + 1. (III)Sn = a1 + a2 + a3 + - + an = (30 + 1) + + (32 + 1) + - + (3n-1 + 1) = (30 + 3 + 32 + - + 3n-1 ) + n =+ n =×3n + n-.

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