Question

二次函数y=x^2-2(n-1)x+n^2-14 记顶点的纵坐标为an,证明数列an为等差数列

Answer 1

y=x^2-2(n-1)x+n^2-14

=[x-(n-1)]^2+n^2-14-n^2+2n-1

=[x-(n-1)]^2+2n-15

顶点：(n-1,2n-15)

an=2n-15

a(n+1)=2(n+1)-15

=2n-13

a(n+1)-an=2n-13-(2n-15)

=2

即公差为2

a1=2*1-15=-13

{an}是首项为-13,公差为2的等差数列.