数列{an}满足a1=1,a(n+1)=2^(n+1)an/an+2^n(n∈N) (1)证明数列{2^n/an}是等差数列,
问题描述:
数列{an}满足a1=1,a(n+1)=2^(n+1)an/an+2^n(n∈N) (1)证明数列{2^n/an}是等差数列,
答
同除以2^(n+1)
得a(n+1)/2^(n+1)=an/(an+2^n)
倒过来得2^(n+1)/a(n+1)=1+[(2^n)/an]
[2^(n+1)/a(n+1)]-[(2^n)/an]=1
得证数列{an}的通项公式怎么求啊?2^n/an=(n-1)+(2/a1)=n+1an=(2^n)/(n+1)