Question

已知函数f（x）满足f（1）=2，f（x+1）=（1+f（x））/（1-f（x）），则f（1）*f（2）*f（3）

则f（1）*f（2）*f（3）……*f（2007）=？
请附上解题过程，O(∩_∩)O谢谢

Answer 1

f(1) = 2
f(2) = f(1+1) = [1+f(1)]/[1-f(1)] = (1+2)/(1-2)
f(2) = -3
f(3) = f(2+1) = [1+f(2)]/[1-f(2)] = [1+(-3)]/[(1-(-3)]
f(3) = -1/2
f(4) = f(3+1) = [1+f(3)]/[1-f(3)] = [1+(-1/2)]/[(1-(-1/2)]
f(4) = 1/3
f(5) = f(4+1) = [1+f(4)]/[1-f(4)] = [1+(1/3)]/[(1-(1/3)]
f(5) = 2

从这里可以看到f(5)=f(1)，也就是说每四次就会重复了
形成了f(1+4x)=2，x≥0

在2005年时，f(2005) = f[1+4(501)]=2
f(2006) = f(2005+1) = [1+f(2005)]/[1-f(2005)] = (1+2)/(1-2)
f(2006) = -3
f(2007) = f(2006+1) = [1+f(2006)]/[1-f(2006)] = [1+(-3)]/[(1-(-3)]
f(2007) = -1/2

而且，当f(1) x f(2) x f(3) x f(4) 时，答案是1
既然每四次就重复，自然的 f(5) x f(6) x f(7) x f(8) 也是1
f(2001) x f(2002) x f(2003) x f(2004) = 1
f(2005) x f(2006) x f(2007)
= 2 x -3 x -1/2
=3

结论 f(1) x f(2) x f(3) x ... ... x f(2007) = 3

纯手打，蒙采纳~