设数列{an}的前n项和为Sn,其中an不等于0.a1为常数,且-a1,Sn,a(n+1)成等差数列,设Bn=1-Sn,问是否存在a1,

存在,an=a1*3^(n-1),Sn=[a(n+1)-a1]/2,Bn=1-[a(n+1)-a1]/2=1-[(3^n)-1]*(a1)/2所以B1=1-a1,B2=1-4a1,B3=1-13a1如果Bn为等比数列,那么有：B1*B3=(B2)^2,即(1-a1)*(1-13a1)=(1-4a1)^2,解得a1=-2代入Bn=1-[(3^n)-1]*(a1...