Question

已知a1=x,an+1=1-1/an(n=1,2,3,…）求a2,a3,a4,a5；求a2002

Answer 1

a2=1-1/a1=1-1/x=(x-1)/x

a3=1-1/a2=-1/(x-1)

a4=1-1/a3=x

a5=1-1/a4=(x-1)/x

综上分析知,an是周期性的,且有a(n)=a(n+3)

则a(2002)=a(667*3+1)=a1=x