已知数列{An}中,A1=1/2,A（n+1）=1-1/An （n≥2）,则A16=?

A1=1/2,A2=-1,A3=2,A4=1/2,A5=-1,由此可见An是周期数列，A4=A(1+3)=A1,An=A(n-3),所以A16=A13=A10=A7=A4=A3=A1=1/2

由题意可知，a1=1/2,a2=-1,a3=2,a4=1/2=a1
由上式可知，{an}是一个每三次循环的数列，所以a15=2，a16=a1=1/2

a(1)=1/2,
a(2) = 1 - 1/a(1) = -1,
a(3) = 1 - 1/a(2) = 2,
a(4) = 1 - 1/a(3) = 1/2 = a(1),
a(5) = 1 - 1/a(4) = 1 - 1/a(1) = a(2),
a(6) = a(3),
...
a(3n-2) = a(1) = 1/2,
a(3n-1) = a(2) = -1,
a(3n) = a(3) = 2,
a(16) = a(3*6-2) = a(1) = 1/2

这是一个循环数列（每三个数循环一次），
A(1)=1/2
A(2)=1-1/A(1)=1-2=-1
A(3)=1-1/A(2)=1+1=2
A(4)=1-1/A(3)=1-1/2=1/2
即A(n)=A(3k+n) n>=1, k>=0
所以A(16)=A(3*5+1)=A(1)=1/2