(1)由框图,知数列{xn}中,x1=1,xn+1=xn+2, ∴xn=1+2(n-1)=2n-1(n∈N*,n≤2008). (2)y1=2,y2=8,y3=26,y4=80. 由此,猜想yn=3n-1(n∈N*,n≤2008). 证明:由框图,知数列{yn}中,yn+1=3yn+2, ∴yn+1+1=3(yn+1),∴=3,y1+1=3, ∴数列{yn+1}是以3为首项,3为公比的等比数列, ∴yn+1=3·3n-1=3n, ∴yn=3n-1(n∈N*,n≤2008). (3)zn=x1y1+x2y2+…+xnyn =1×(3-1)+3×(32-1)+…+(2n-1)(3n-1) =1×3+3×32+…+(2n-1)·3n-[1+3+…+(2n-1)] 记Sn=1×3+3×32+…+(2n-1)·3n ① 则3Sn=1×32+3×33+…+(2n-1)·3n+1 ② ①-②,得-2Sn=3+2·32+2·33+…+2·3n-(2n-1)·3n+1 =2(3+32+…+3n)-3-(2n-1)·3n+1 =2×-3-(2n-1)·3n+1 =3n+1-6-(2n-1)·3n+1 =2(1-n)·3n+1-6, ∴Sn=(n-1)·3n+1+3. 又1+3+…+(2n-1)=n2, ∴zn=(n-1)·3n+1+3-n2(n∈N*,n≤2008). |