a1=3>0 又a(n+1)=an²>0,数列各项都>0
a(n+1)=an²
lga(n+1)=lg(an²)=2lgan
lga(n+1)/lgan=2,为定值.
lga1=lg3
数列{lgan}是以lg3为首项,2为公比的等比数列.
lgan=2^(n-1)×lg3=lg[3^(2^(n-1))]
an=3^[2^(n-1)]
数列{an}的通项公式为an=3^[2^(n-1)].