证明;可设通项an=a1×q^(n-1).n=1,2,3,[[[1]]]当q=1时,易知an=a1,Sn=na1,n=1,2,3.由题设2S9=S3+S6即18a1=3a1+6a1a1=0.矛盾.∴q≠1[[[2]]]当q≠1时.易知Sn=[a1/(q-1)]×[-1+q^n].n=1,2,3,...由题设2S9=S3+S6可得2[-1+...