已知数列{an}满足a1=1,a2=2,a(n+2)=an+a(n+1）,n∈N*(1)令bn=a(n+1)-an,证明{bn}是等比数列

题目

已知数列{an}满足a1=1,a2=2,a(n+2)=an+a(n+1）,n∈N*(1)令bn=a(n+1)-an,证明{bn}是等比数列
（2）求{an}的通项公式,
a(n+2)=【an+a(n+1）】/2

答案

（1）证明：2a(n+2)=an+a(n+1)∴2[a(n+2)-a(n+1)]=an-a(n+1)=-[a(n+1)-an]bn=a(n+1)-an,∴2b(n+1)=-bn,即b(n+1)/bn=-1/2∴{bn}是等比数列（2）2a(n+2)=an+a(n+1)等式俩边同时减去2a(n+1)∴2[a(n+2)-a(n+1)]=an-a(n+1...