已知数列{a[n]}满足a[1]=2，a[n+1]=(1+a[n])/(1-a[n])，求通项。
解：∵a[n+1]=(1+a[n])/(1-a[n])
∴令a[n]=tanθ，则a[n+1]=[tan(π/4)+tanθ]/[1-tan(π/4)tanθ]=tan(π/4+θ)
∵θ=arctan(a[n])，π/4+θ=arctan(a[n+1])
∴上面两式相减，得：arctan(a[n+1])-arctan(a[n])=π/4
∵a[1]=2
∴{arctan(a[n])}是首项为arctan(a[1])=arctan2，公差为π/4的等差数列
即：arctan(a[n])=arctan2+(n-1)π/4
∴a[n]=tan[(n-1)π/4+arctan2]

其实可以用不动点，
lz
苦劳是有滴