已知数列An的前N项和为Sn，且An=1,A(n+1)=1/3Sn，求a2,a3,a4的值及数列An的通项公式

a(2) = s(1)/3 = a(1)/3 = 1/3.
a(3) = s(2)/3 = [a(1) + a(2)]/3 = [1 + 1/3]/3 = 4/9
a(4) = s(3)/3 = [a(1) + a(2) + a(3)]/3 = [1 + 1/3 + 4/9]/3 = 16/27.

a(n+1) = s(n)/3,
a(n) = s(n-1)/3,

a(n+1) - a(n) = [s(n) - s(n-1)]/3 = a(n)/3,
a(n+1) = a(n)[4/3].

{a(n)} 是首项为1，公比为4/3的等比数列，
a(n) = (4/3)^(n-1), n = 1,2,...

an=(1/3)*(4/3)^n-2