摘要: 设数列{an}的首项a1=1.前n项和Sn满足关系式:3tSn-(2t+3)Sn-1=3t(t>0,n=2,3,4,-). (Ⅰ)求证:数列{an}是等比数列, (Ⅱ)设数列{an}的公比为f(t).作数列{bn}使b1=1,bn=f( )(n=2,3,4,-).求bn; (Ⅲ)求和:b1b2-b2b3+b3b4-b4b5+-+b2n-1b2n-b2nb2n+1.

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