设数列{an}满足a1=0,2an+1=1+anan+1 (1)求证:为等差数列,并求通项公式an (2)记 ,求数列{bn}的前n
问题描述:
设数列{an}满足a1=0,2an+1=1+anan+1 (1)求证:为等差数列,并求通项公式an (2)记 ,求数列{bn}的前n
答
2a(n+1)=1+a(n)a(n+1)移向整理得到2a(n+1)-a(n)a(n+1)=1,即[2-a(n)]a(n+1)=1都除以2-a(n)得到a(n+1)=1/[2-a(n)]设c(n)=a(n)/[a(n)-1]所以c(n+1)=a(n+1)/[a(n+1)-1]=1/[2-a(n)]/{[1/[2-a(n)]-1}=1/[a(n)-1]所以c(n+1)...