设{an}的公比为q,则q>0
∵,a1=3,a3=a1q²=243
∴q²=81 ,q=9
∴an=3*9^(n-1)=3^(2n-1)
∴bn=log₃an=log₃3^(2n-1)=2n-1
∴1/[bnb(n+1)]=1/[(2n-1)(2n+1)]=1/2[1/(2n-1)-1/(2n+1)]
∴Sn=1/2[1-1/3+1/3-1/5+.+1/(2n-1)-1/(2n+1)]
=1/2[1-1/(2n+1)]
=n/(2n+1)