Question

一道高三数学题，求解

Answer 1

导函数f'(x)= x^2+ax+b是偶函数,则有a=0
又有f'(1)=2,得到b=1
即有f'(x)=x^2+1
a(n+1)=2f'(an-1)-1=2(an-1)^2+2-1
a(n+1)-1=2(an-1)^2
a1=3, b1=log2(a1-1)=1

(b(n+1)+1)/(bn+1)=[log2(a(n+1)-1)+1]/(log2(an-1)+1)=[log2(2(an-1)^2)+1]/(log2(an-1)+1)
=[1+2log2(an-1)+1]/(log2(an-1)+1)
=2
故有数列{bn+1}是一个首项是2,公比是2的等比数列.

Answer 2

Answer 3

追问

谢谢了！

Answer 4

Answer 5

f'(x)=x^2+ax+b是偶函数a=0
f'(x)=x^2+b，f'(1)=1+b=2，b=1
f'(x)=x^2+1
a(n+1)=2(an-1)^2+1-1
a(n+1)=2(an-1)^2
log(2,a(n+1))=1+2log(2,an-1)
b(n+1)=1+2bn
b(n+1)+1=2(bn+1)
得证。