Question

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

Answer 1

a1=f(x+1)=(x+1)^2-4(x+1)+2=x^2-2x-1

a2=f(x-1)=(x-1)^2-4(x-1)+2=x^2-6x+7

d=a2-a1=8-4x

a(n)=a1+(n-1)d=x^2-2x-1+(n-1)(8-4x)

==4(2-x)(n-1)+x^2+2x-9