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當n=1時,a1=S1=1+3+1=5;當n≥2時,an=Sn-Sn-1=n2+3n+1-[(n-1)2+3(n-1)+1]=2n+2.∴數列{an}的通項公式為an=5,n=12n+2,n≥2.故答案為an=5,n=12n+2,n≥2.
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