摘要：[解] (1)∵an= ∴An=[Cn1(1-q)+Cn2(1-q2)+-+Cnn(1-qn)] =[ Cn1+ Cn2+-+ Cnn-( Cn1q+ Cn2q+-+ Cn1qn)] =[(2n-1)-(1+q)n+1]= [2n-(1+q)n] (2)=[1-()n] ∵-3<q<1,∴||<1 ∴=

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