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).

(1)求通项an.(2)求a2+a5+a8+…+a26

Answer 1

f(x+1)=x^2-2x=(x+1)^2-4(x+1)+3
f(x)=x^2-4x+3
f(x-1)=x^2-6x+8
a1=f(x-1)=x^2-6x+8
a2=-1/2
a3=x^2-4x+3
等差数列{an}中
a1+a3=2a2
2x^2-10x+11=-1
x^2-5x+6=0
x=2或x=3
1）若x=2，a1=0，a3=-1
数列{an}是首项为0，公差为-1/2
a2+a5+a8+…+a26=(a2+a26)*9/2=-117/2
2)若x=3，a1=-1，a3=0
数列{an}是首项为-1，公差为1/2
a2+a5+a8+…+a26=(a2+a26)*9/2=99/2 f(x+1)=x^2-2x,令x+1=t
f(t)=(t-1)^2-2(t-1)=t^2-4t+3
a1=f(x-1)=(x-1)^2-4(x-1)+3=x^2-6x+8
a2-a1=a3-a2
-1/2-(x^2-6x+8)=x^2-4x+3+1/2
x^2-5x+6=0
x=2或x=3
(1)x=2
a1=0
q=a2-a1=-1/2
a2+a5+a8+…+a26＝n*a2+n*(n-1)(3q)=9*(-1/2)+9*(9-1)(-3*1/2)
=-225/2
(2)x=3
a1=-1
q=a2-a1=1/2
a2+a5+a8+…+a26＝n*a2+n*(n-1)(3q)=9*(-1/2)+9*(9-1)(3*1/2)
=207/2