题目

已知数列{an}满足a1=1,an=a1+1/2a2+1/3a3+…+1/(n-1)a(n-1),若an=2006,则n=___

提问时间：2020-11-03

答案

a2=a1=1n>=3时an+1=a1+1/2a2+.+1/n-1an-1+1/nan两式相减得an+1-an=1/nan即an+1=n+1/nan即an+1/an=n+1/nan+1/a2=(an+1/an)(an/an-1).(a3/a2)=(n+1/n)(n/n-1).(3/2)=n+1/2即an+1=n+1/2即an=n/2an=2006,n=4012...