Question

已知数列{an}的各项都是正数且满足a0=1,an+1=an(4-an)/2(n∈N），求数列{an}的通项公式

Answer 1

a(n+1)=(1/2)an(4-an)
2a(n+1)=4an-an^2
=-[an^2-2*2an+4]+4
=-(an-2)^2+4
2[a(n+1)-2]=-(an-2)^2
设bn=an-2,b0=a0-2=-1
2b(n+1)=-(bn)^2
b(n+1)=(-1/2)(bn)^2
=(-1/2){(-1/2)[b(n-1)]^2}^2=(-1/2)^3*[b(n-1)]^4
=(-1/2)^3*{(-1/2)[b(n-2)]^2}^4=(-1/2)^7*[b(n-2)]^8
……
=(-1/2)^[2^(n-1)-1]*(b2)^[2^(n-1)]
=(-1/2)^[2^n-1]*(b1)^[2^n]
=(-1/2)^[2^(n+1)-1]*(b0)^[2^(n+1)]
=(-1/2)^[2^(n+1)-1]*(-1)^[2^(n+1)]
=(-1/2)^[2^(n+1)-1]
bn=(-1/2)^[2^n-1]

an=2+bn=2+(-1/2)^[2^n-1]
a(n+1)=2+(-1/2)^[2^(n+1)-1]＜2
a(n+1)-an=(-1/2)^[2^(n+1)-1]-(-1/2)^[2^n-1]
={(-1/2)^[2^n-1]}{(-1/2)^[2^(n+1)-2^n]-1}
=[(-1/2)^(2^n-1)]{(-1/2)^(2^n)-1}
又因2^n-1为奇数，所以(-1/2)^(2^n-1)＜0；
因0＜(-1/2)^(2^n)＜1为奇数，所以(-1/2)^(2^n)-1＜0
所以[(-1/2)^(2^n-1)]{(-1/2)^(2^n)-1}＞0
所以a(n+1)-an＞0，an＜a(n+1)，

综上所述an＜a(n+1)＜2。