等比数列{an}的前n项和为sn,满足sn=1/3^n+a，n=1时a1=1/3+a,n>1时an=Sn-S=1/3^n-1/3^(n-1)=-2/3^n,∴公比q=a/an=1/3=a2/a1=(-2/9)/(1/3+a),∴1/3+a=-2/3,a=-1.∴an=-2/3^n(n∈N+),an^2=4/9^n,∴{an^2}的前n项和=[4/9-4/9^(n+1)]/(1-1/9)=(1/2)(1-1/9^n).
