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解析 通項公式an=sn-s(n-1) 所以 an=n²;+n+1-(n-1)²;-(n-1)-1 =n²;+n-(n²;-2n+1)-n+1 =n²;-n²;+2n-1+1 =2n 當n=1時 a1=3 所以an的通項公式{a1=3 n=1 an=2n n≥2
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