f(x,n)=(1＋x/2)(1＋x/2^2)…(1＋x/2^n)=g(x,n)x³+b(n)x²+a(n)x+1
f(x,n+1)=f(x,n)(1＋x/2^(n+1))
g(x,n+1)x³+b(n+1)x²+a(n+1)x+1=[g(x,n)x³+b(n)x²+a(n)x+1][1＋x/2^(n+1)]
比较x系数有 a(n+1)=a(n)+1/2^(n+1)
比较x²系数有b(n+1)=b(n)+a(n)/2^(n+1)
f(x,1)=1+x/2 知a(1)=1/2 b(1)=0
所以a(n)=a(n-1)+1/2^n=a(n-2)+1/2^(n-1)+1/2^n=.=a(1)+1/2^2+1/2^3+.+ 1/2^(n-1)+1/2^n
=1/2+1/2^2+.+1/2^n=1-1/2^n
b(n+1)=b(n)+a(n)/2^(n+1)=b(n)+1/2^(n+1)-1/2^(2n+1)
所以b(n)=b(n-1)+ 1/2^n-1/2^(2n-1) =b(n-2)+1/2^(n-1)+1/2^n-1/2^(2n-3)-1/2^(2n-1)
=b(1)+(1/2^2+1/2^3+.+1/2^n)-(1/2^3+1/2^5+.+1/2^(2n-1))
=0+(1/2-1/2^n)-(1/6)[1-1/2^(2n-2)]
=1/3-1/2^n+1/[3×2^(2n-1)]
a(n)=1-(1/2)^n b(n)=1/3-1/2^n+1/[3×2^(2n-1)]
(2) 是否存在常数p.q（p