Question

有一列数a1，a2，a3，...，aN从第二个数开始，每一个数都等于1与它前面那个数的倒数的差，若a1=2，则a2012

等于几啊，，，，要算式

Answer 1

a[n] = 1 - 1/a[n-1]= (a[n-1]-1)/a[n-1]
a[n+1]
= 1 - 1/a[n]
= 1 - a[n-1]/(a[n-1]-1)
= - 1/(a[n-1]-1)
a[n+2] = 1 - 1/a(n+1)=1- [1-a(n-1)]=a[n-1]
∴a[n+2]=a[n-1] 周期为3的函数
2012/3=670余2
∴ a[2012] = a[2] = 1 - 1/a[1] = 1-1/2=1/2