Question

数列{an}为等差数列,a1=f（x+1）,a2=0,a3=f（x-1）,其中f（x）=x平方--4x+2.求通项公式a

Answer 1

等差数列:a2-a1=a3-a2 a2=0所以:

a1+a3=0

将f（x+1）和f（x-1）展开代入a1+a3=0,有:

x^2-4x+3=0,所以x=1或者x=3

x=1时,代入a1=f（x+1）得a1=-2,a2-a1=2=d

此时:an=-2+2(n-1)

x=3时,代入a1=f（x+1）得a1=2,a2-a1=-2=d

此时:an=2-2(n-1)