见解析
解:(1)由程序框图可知:{xn}是等差数列,且首项x1=1,公差d=2
∴xn=1+2(n-1)=2n-1…(3分)
(2)y1=2=3-1,y2=3×2+2=8=32-1,y3=3×8+2=26=33-1,y4=3×26+2=80=34-1
故yn=3n-1…(7分)
(3)xn?yn=(2n-1)(3n-1)=(2n-1)?3n-(2n-1),
∴zn=(3-1)+(3?32-3)+(5?33-5)+…+[(2n-1)?3n-(2n-1)]
=3+3?32+5?33+…+(2n-1)?3n-[1+3+5+(2n-1)]
=3+3?32+5?33+…+(2n-1)?3n-n2
令Sn=3+3?32+5?33+…+(2n-1)?3n3Sn
=32+3?33+5?34+…+(2n-1)?3n+1
∴-2Sn=3+2(32+33+34+…+3n)-(2n-1)?3n+1
=2(1-n)?3n+1-6
∴Sn=(n-1)?3n+1+3
∴zn=(n-1)???3n+1+3-n2…(14分)