以知数列{an},{bn}满足条件a1=b1=1,且an=a(n-1)+2b(n-1),n=}的通项公式an=?以知数列{an},{bn}满足条件a1=b1=1,且an=a(n-1)+2b(n-1),bn=2a(n-1)-b(n-1),n=2,3,4,...则{an}的通项公式an=?
问题描述:
以知数列{an},{bn}满足条件a1=b1=1,且an=a(n-1)+2b(n-1),n=}的通项公式an=?
以知数列{an},{bn}满足条件a1=b1=1,且an=a(n-1)+2b(n-1),bn=2a(n-1)-b(n-1),n=2,3,4,...则{an}的通项公式an=?
答
a1=1
a2=3
a(n)=5a(n-2) n>2
答
由已知得:an-a(n-1)=2b(n-1).①a(n+1)-an=2bn.②①+②,得:a(n+1)-a(n-1)=2[bn+b(n-1)]=4a(n-1)得a(n+1)=5a(n-1)即a(n+2)=5an所以数列{an}隔项成等比又因为a1=b1=1,a2=a1+2b1=1+2=3所以有:an=5^[(n-1)/2](n为奇数)=...