已知数列｛an｝满足a1=2,a(n+1)=1+an/1-an（n∈N*）,则a2014的值为

特征方程为
r = (1+r)/(1-r),
r - r^2 = 1+r,
0 = 1+r^2.
特征方程没有实数根,说明数列为周期数列.
a(1)=2,
a(2) = [1+a(1)]/[1-a(1)] = 3/(-1) = -3.
a(3) = [1+a(2)]/[1-a(2)] = (-2)/(4) = -1/2.
a(4) = [1+a(3)]/[1-a(3)] = (1/2)/(3/2) = 1/3.
a(5) = [1+a(4)]/[1-a(4)] = (4/3)/(2/3) = 2 = a(1),
a(6) = a(2),
...
a(4n-3)= a(1) = 2,
a(4n-2) = a(2) = -3,
a(4n-1) = a(3) = -1/2,
a(4n) = a(4) = 1/3.
a(2014) = a(4*504-2) = a(2) = -3.