已知各项为正数公比q的等比数列{an}中，Sn为它的前n项和，a3=1/9,S2=4/3

(1) S2 = a1+a1*q = 4/3, a3 = a1*q^2 = 1/9
两式相除得到 (1+q)/q^2 = 12
又由正项等比数列q>0，解得q=1/3
(2) 由上题还可解得a1=1
an = (1/3)^(n-1)
bn = log1/9 an = (n-1)/2
bn是以0为首项，1/2为公差的等差数列
S8 = (b1+b8)*4 = 7/2*4 = 14

a(n) = aq^(n-1), a > 0, q > 0.
1/9 = a(3) = aq^2,
4/3 = S(2) = a + aq = a(1+q).
(1/9)*3/4 = q^2/(1+q) = 1/12,
0 = 12q^2 - q - 1 = (4q+1)(3q-1), q = 1/3.
a = 1/(9q^2) = 1.
a(n) = (1/3)^(n-1) = 1/3^(n-1), n = 1,2,...

b(n) = log(1/9) a(n) = ln[(1/3)^(n-1)]/ln(1/9) = (n-1)ln(1/3)/[2ln(1/3)] = (n-1)/2, n = 1,2,...
B(n) = b(1)+b(2)+...+b(n) = (1/2)n(n-1)/2 = n(n-1)/4,
B(8) = 8*7/4 = 14