等比数列{an}的前n项和为Sn,已知S1,S3,S2成等差数列

S1=a1S2=a1(1+q)S3=a1(1+q+q^2)S1,S3,S2成等差数列即s3-s1=s2-s31+q+q^2-1=1+q-(1+q+q^2)q^2+q=-q^2q=0或-1/2如果a1-a3=3a1不等于a3q不等于0,即q=-1/2a1(1-1/4)=3a1=4所以Sn=4*(1-(-1/2)^n)/(3/2)=8*(1-(-1/2)^n)/3...