S1/a1=1S2/a2-S1/a1=(2+d)/(1+d)-1=d/(1+d)S3/a3-S1/a1==(3+3d)/(1+2d)-1=(2+d)/(1+2d)2*d/(1+d)=(2+d)/(1+2d)解得d=1,d=0(舍去)所以,an=n (b(n+1))^2=q^(n+1)(n+2)=q^(n*n+3n+2)2*bn*b(n+2)=2*q^{n(n+1)/2+(n+1)(n+2)/2}=2*q^(n*n+3n+3)