an=1+1/3+1/3²+1/3(n-1)=[1-(1/3)^n]/(1-1/3)=3/2-1/[2*3^(n-1)]
前n项和sn=(3/2)-1/[2*1]+(3/2)-1/[2*3^1]+.+3/2-1/[2*3^(n-1)]
=3n/2-(1/2)[1+1/3+...+1/3^(n-1)]
=3n/2-(1/2)[1-1/(3^n)]/(1-1/3)
=3n/2-(3/4)*[1-1/(3^n)]
=3n/2-3/4+1/[4*3^(n-1)]