Question

设数列<an＞满足:a1=3,an+1=1+an/1-an (n＞=1)则a2014=多少

设数列<an＞满足:a1=3,an+1=1+an/1-an (n＞=1)则a2014=多少 注:an+1的n+1为角标

Answer 1

a(n+1) = [1+a(n)]/[1-a(n)],
特征方程为：
r = (1+r)/(1-r)， r - r^2 = 1 + r, 0 = 1+r^2.无实数解。
因此，数列必为周期数列。

a(1) = 3,
a(2) = [1 + a(1)]/[1-a(1)] = (1+3)/(1-3) = -2.
a(3) = [1+a(2)]/[1-a(2)] = (1-2)/(1+2) = -1/3.
a(4) = [1+a(3)]/[1-a(3)] = [1-1/3]/[1+1/3] = 1/2.
a(5) = [1+a(4)]/[1-a(4)] = [1+1/2]/[1-1/2]= 3 = a(1).
a(6) = [1+a(5)]/[1-a(5)] = [1+a(1)]/[1-a(1)] = a(2).
a(7) = a(3),
a(8) = a(4).
...
a(4n-3) = a(1) = 3,
a(4n-2) = a(2) = -2,
a(4n-3) = a(3) = -1/3.
a(4n) = a(4) = 1/2.

a(2014) = a(4*504 - 2) = a(2) = -2.

追问

哦 我算到第4个发现又不是等差，又不是等比，就不想算了，，原来有规律的。。谢了

Answer 2

1a=3,a2=-1/2,a3=2/3,a4=4,a5=-1/2,循环了a2014=a1=3