若公比q=1S6/S3=6a1/(3a1)=2,与已知不符,因此公比q≠1S6/S3=4[a1(q^6-1)/(q-1)]/[a1(q^3-1)/(q-1)]=4(q^6-1)/(q^3-1)=4(q^3+1)(q^3-1)/(q^3-1)=4q^3+1=4q^3=3q=3^(1/3)an=a1q^(n-1)=1×3^[(n-1)/3]=3^[(n-1)/3]...