在数列｛an｝中,a1=1,Sn=(3an)－1（n≥2）,求通项公式

a1=1
n≥2,
an=Sn-S(n-1)=3an-1-[3a(n-1)-1]=3an-3a(n-1)
2an=3a(n-1)
an/a(n-1)=3/2
所以an=a1x(3/2)^(n-1)

s(n+1) = 3a(n+1) - 1, s(2) = a(1)+a(2) = 3a(2) - 1, a(2)= [1+a(1)]/2 = 1.
s(n+2) = 3a(n+2) - 1,
a(n+2) = s(n+2)-s(n+1) = 3a(n+2) - 3a(n+1),
a(n+2) = (3/2)a(n+1),
{a(n+1)}是首项为a(2)=1,公比为3/2的等比数列。
a(n+1) = (3/2)^(n-1),
a(1)=1,
n>=2时，a(n) = (3/2)^(n-2)