Question

已知f(x+1)=x^2-2x,等差数列{an}中,a1=f(x-1),a2=-1/2,a3=-f(x),求an及a2+

Answer 1

f(x+1)=x²-2x=(x-1)²-1=[(x+1)-2]²-1=(x+1)²-4(x+1)+3

f(x)=x²-4x+3

a1=f(x-1)=(x-1)²-4(x-1)+3=x²-6x+8

a2=-1/2

a3=-f(x)=-x²+4x-3

a1,a2,a3成等差数列,则

2a2=a1+a3

2(-1/2)=x²-6x+8-x²+4x-3

整理,得

2x=6

x=3

代入a1表达式,a1=x²-6x+8=9-18+8=-1

d=a2-a1=(-1/2)-(-1)=1/2

数列{an}是以-1为首项,1/2为公差的等差数列.

an=-1+(1/2)(n-1)=(n-3)/2

a2+a4+...+a20

=(2+4+...+20)/2-30/2

=55-15

=40