Question

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

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.