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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解：
∵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

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解：(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)。

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q<2讨论） bn=1 or 4*(-2)^(n-1)

1)

由题意可得a1=[(d-1)-1]²=(d-2)²=d²-4d+4

a3=[(d+1)-1]²=d²

因为{an}是公差为d的等差数列

所以a3-a1=(a1+2d)-a1=2d

所以d²-(d²-4d+4)=2d ==>d=2

b1=[(q+1)-1]²=q²

b3=[(q-1)-1]²=(q-2)²=q²-4q+4

因为{bn}是公比为q的等比数列 因为b3=b1q²

所以b3/b1=q²

地铺哦(q²-4q+4)/q²=q²

==>q^4-(q²-4q+4)=0

==>(q²)²-(q-2)²=0

==>(q²-q+2)(q²+q-2)=0

因为q²-q+2=(q-1/2)²+7/4>0

所以只有q²+q-2=0 ==>q=1或-2

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