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+a4+...+a20的值

由题意可知：f(x) = (x-1)^2 -2(x-1) = x^2-4x +3
故a1 = f(x-1) = (x-1)^2 - 4(x-1) + 3 = x^2 - 6x +8
a3 = -f(x) = -x^2 + 4x - 3
因为数列{an}是等差数列所以2a2 = a1 + a3
-1 = x^2 - 6x +8 + (-x^2 + 4x - 3) 故 x = 3，所以a1 = -1, 公差d = a2 - a1 = 1/2
那么an = a1 + (n - 1)d = -1 + (n - 1) 1/2 = n/2 - 3/2
a4 - a2 = 1,以a2为首项，公差为1的等差数列的前十项和为10(a2 + a20)/2 = 40,即为所求

是的

f(x+1)=x²-2x=(x-1)²-1=[(x+1)-2]²-1=(x+1)²-4(x+1)+3f(x)=x²-4x+3a1=f(x-1)=(x-1)²-4(x-1)+3=x²-6x+8a2=-1/2a3=-f(x)=-x²+4x-3a1,a2,a3成等差数列,则2a2=a1+a32(-1/2)=x...