a(n) = 8 + (n-1)d.
a(1)b(1) = 8b(1) = 1*2^(1+3) = 16,b(1) = 2.
b(n) = 2q^(n-1).
a(1)b(1)+a(2)b(2)+...+a(n)b(n) = n*2^(n+3),
a(1)b(1)+a(2)b(2)+...+a(n)b(n)+a(n+1)b(n+1) = (n+1)*2^(n+4),
a(n+1)b(n+1) = (n+1)2^(n+4) - n*2^(n+3) = (2n+2-n)2^(n+3) = (n+2)2^(n+3).
a(n)b(n) = (n+1)2^(n+2).
(n+1)2^(n+2) = [8+(n-1)d]*2q^(n-1),
(n+1)2^(n+1) = [8+(n-1)d]q^(n-1).
n=2时,3*2^3 = [8 + d]q,24 = (8+d)q.
d不为-8.
q = 24/(8+d).
n=3时,4*2^4 = [8+2d]*q^2,
32 = (4+d)q^2 = (4+d)*[24/(8+d)]^2,
32(8+d)^2 = 16*36(4+d),
(8+d)^2 = 18(4+d),
0 = d^2 + 16d + 64 - 18d - 72 = d^2 - 2d - 8 = (d-4)(d+2),
d=4时,q = 24/(8+d) = 2.
a(n) = 8 + 4(n-1) = 4n+4.
b(n) = 2q^(n-1) = 2^n.
a(n)b(n) = 4(n+1)2^n = (n+1)2^(n+2),满足题意.
d=-2时,q=24/(8+d) = 4.
a(n) = 8 - 2(n-1)= 10 - 2n.
b(n) = 2q^(n-1) = 2*4^(n-1) = 2^(2n-1).
a(n)b(n) = (10-2n)2^(2n-1)与a(n)b(n)=(n+1)2^(n+2)矛盾.
因此,只有,
a(n) = 4n+4,
b(n) = 2^n.
第2部分,题意不清.
