已知数列{an}的各项都是正数,且满足a0=1,an+1(n+1是a的角标)=1/2an(4-an)证明an

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  • 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.