设数列{an}是公比为q的等比数列，如果S5-S3=4/81，a2+a4=4/9，那么S6=？

a(n)=aq^(n-1),n=1,2,...
4/9 = a(2)+a(4)=aq+aq^3=aq(1+q^2),
若q=1,则4/9=2a, a = 2/9.
a(n)=2/9,
S(n)=2n/9,
S(5)-S(3)=2/9[5-3]=4/9不等于4/81.
因此，q不等于1.
S(n)=a[q^n-1]/(q-1).
4/81=S(5)-S(3)=[a/(q-1)][q^4-q^2]=a[q^2+1][q+1],
1/9 = (4/81)/(4/9)=a[q^2+1][q+1]/[aq(q^2+1)] = (q+1)/q,
q = 9q+9,
q=-9/8.
4/9=aq(1+q^2)=-9a/8*(1+81/64), a = -4*8/[81*(1+81/64)] = -32*64/[81*145] = -2048/11745,
S(6)=a[q^6-1]/(q-1)=-32*64[1 - (9/8)^6]/[81*145*(1+9/8)]
= 4*8^4[(9/8)^6-1]/[81*145*17]
= [9^6 - 8^6]/[32*81*5*29*17]
= [9^3+8^3][9^3-8^3]/[32*81*5*29*17]
= [9+8][9^2+8^2-8*9][9-8][9^2+8^2+8*9]/[32*81*5*29*17]
= [81+64-72][81+64+72]/[32*81*5*29]
= [73*217]/[32*81*5*29]
= [7*31*73]/[32*81*5*29]