Question

数列{an}是公差为正数的等差数列,a1=f(x-1),a2=0,a3=f(x+1),其中f(x)=x^2-4x+2,则

Answer 1

设公差为d

a1=f(x-1)=(x-1)^2-4(x-1)+2=x^2-6x+7

a3=f(x+1)=(x+1)^2-4(x+1)+2=x^2-2x-1

a1+a3=2a2

x^2-6x+7+x^2-2x-1=0

x^2-4x+3=0

x1=1,x2=3

x=1时,a1=2,a3=-2,d=(a3-a1)/2=-20

所以an=a1+(n-1)d=-2+2(n-1)=2n-4

所以an=2n-4