已知等比数列{an}中,a2a3a4分别是某等差数列的第5项第3项第2项，且a1=1⼀2,公比q≠1

a(n)=(1/2)q^(n-1),
a(2)=q/2=c(5)=c+4d,
a(3)=q^2/2=c(3)=c+2d,
a(4)=q^3/2=c(2)=c+d,

q/2 - q^3/2 = (c+4d)-(c+d)=3d,
q^2/2 - q^3/2 = (c+2d)-(c+d)=d,
q/2 - q^3/2 = 3d = 3[q^2/2 - q^3/2],
q - q^3 = 3[q^2 - q^3],
0 = 2q^3 - 3q^2 + q = q[2q^2 - 3q + 1] = q(2q-1)(q-1), q=1/2.
a(n)=(1/2)(1/2)^(n-1)=1/2^n,

a(1)b(1)=2-1=1=b(1)/2, b(1)=2.
2(n+1)-1 - (2n-1) = 2 = a(n+1)b(n+1) = b(n+1)/2^(n+1),
b(n+1) = 2^(n+2) = 2^(n+1+1),
b(1)=2,
n>=2时，b(n) = 2^(n+1),

s(1)=b(1)=2,
n>=2时，s(n)=2+2^3 + 2^4 + ... + 2^n + 2^(n+1)=2+2^2+2^3 + ... + 2^(n+1) - 2^2
=2[1+2+...+2^n] - 4
=2[2^(n+1)-1]/(2-1) - 4
=2[2^(n+1)-1] - 4
=2^(n+2) - 6,

综合，有
s(n) = 2^(n+2) - 6