一道数列问题,通项公式不知道怎么求

3个回答

  • 周期数列.

    a(n+1)=[1+a(n)]/[1-a(n)],

    a(n)不为1.

    r = (1+r)/(1-r),r - r^2 = 1 + r,0 = 1+r^2.特征方程无实数解,因此为周期数列.

    a(2) = [1 + a(1)]/[1-a(1)],

    1 + a(2) = [ 1 + a(1)]/[1 - a(1)] + 1 = [1 + a(1) + 1-a(1)]/[1-a(1)] = 2/[1-a(1)],

    1 - a(2) = 1 - [1+a(1)]/[1-a(1)] = [1-a(1)-1-a(1)]/[1-a(1)] = -2a(1)/[1-a(1)].

    a(3) = [1+a(2)]/[1-a(2)] = {2/[1-a(1)]}/ [-2a(1)/[1-a(1)]} = -1/a(1),

    1 + a(3) = 1 - 1/a(1) = [a(1)-1]/a(1),

    1 - a(3) = 1 + 1/a(1) = [a(1)+1]/a(1),

    a(4)= [1+a(3)]/[1-a(3)] = [a(1)-1]/[a(1)+1],

    1 + a(4) = [a(1)+1+a(1)-1]/[a(1)+1] = 2a(1)/[a(1)+1],

    1 - a(4) = [a(1)+1-a(1)+1]/[a(1)+1] = 2/[a(1)+1],

    a(5)=[1+a(4)]/[1-a(4)] = 2a(1)/2 = a(1),

    a(6)=[1+a(5)]/[1-a(5)] = [1+a(1)]/[1-a(1) = a(2),

    ...

    a(4n-3)=a(1),

    a(4n-2)=a(2) = [1+a(1)]/[1-a(1)],

    a(4n-1)=a(3) = -1/a(1),

    a(4n) = [a(1)-1]/[a(1)+1],

    a(1)不为0,a(1)不为1,a(1)不为-1.