答案：当i=1时z1=n+1,当i!=1时 z=1+[i(1-i^n)/(1-i)],故z2=1+[2(1-2^n)/(1-2)]=2*2^n-1,所以z1*z2=(n+1)*(2*2^n-1)=2(n+1)*2^n-n-1.
所以B={w|w=2(n+1)*2^n-n-1,n属于N}