数列{An}{Bn}满足下列条件：A1=0,A2=1,An+2=An+An+1/2,Bn=An+1-An
问题描述：
数列{An}{Bn}满足下列条件：A1=0,A2=1,An+2=An+An+1/2,Bn=An+1-An
1.求证{Bn}是等比数列 2.求{Bn}的通项公式
答
1.a(n+2)=[an+a(n+1)]/22a(n+2)=an+a(n+1)2a(n+2)-2a(n+1)=an-a(n+1)2[a(n+2)-a(n+1)]=-[a(n+1)-an][a(n+2)-a(n+1)]/[a(n+1)-an]=-1/2,为定值.a2-a1=1-0=1,数列{a(n+1)-an}是以1为首项,-1/2为公比的等比数列.bn=a(n+...