已知f(x)=(x-1)^2,数列{an}是公差为d的等差数列,{bn}是公比为q的等比数列,设a1=f(d-1）已知f(x)=(x-1)^2,数列{an}是公差为d的等差数列,{bn}是公比为q的等比数列,设a1=f(d-1),a3=f(d+1),b1=f(q+1),b3=f(q-1)求{an}{bn}的通项公式

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a1=(d-2)^2=d^2-4d+4 , a3= d^2
a3-a1 = 2d = 4d-4
==> d=2 , a1=0
an=a1+n*d = 2n
b1=q^2 , b3= (q-2)^2 = q^2-4q+4
b3/b1 = q^2 = (q^2-4q+4)/q^2
==>q=1 or -2，b1=1 or 4（分q>2和qbn=1 or 4*(-2)^(n-1)

∵a1=f(d－1)=(d－2)2,a3=f(d+1)=d2,
∴a3－a1=d2－(d－2)2=2d,
∴d=2,∴an=a1+(n－1)d=2(n－1)；又b1=f(q+1)=q2,b3=f(q－1)=(q－2)2,
∴=q2,由q∈R,且q≠1,得q=－2,
∴bn=b*qn－1=4*(－2)n－1

(1)、a1=f(d-1)=(d-2)^2=d^2-4d＋4，a3=f(d＋1)=d^2，又等差数列{an}公差为d，则a3=a1＋2d，即d^2=d^2-2d＋4，解得d=2，有a1=0，则an=a1＋(n-1)d=2n-2 (n=1,2,3...)；(2)、b1=q^2，b3=(q-2)^2，由b3=b1＊q^2得(q-2)^2=q^4，解得：q=1或q=-2，从而bn=1或bn=(-2)^(n＋1)。