在数列{an}中,a1=1,a2=2,a(n+2)=2/3a(n+1)+1/3an,求an+2)=2/3a(n+1)+1/3an,求an
问题描述:
在数列{an}中,a1=1,a2=2,a(n+2)=2/3a(n+1)+1/3an,求an+2)=2/3a(n+1)+1/3an,求an
答
a(n+2)=2/3a(n+1)+1/3an.a3=2/3a2+1/3a1全部相加左边=a3+..a(n+2)右边=2/3a(n+1)+an+.+a3+a2+1/3a1相消后a(n+2)+1/3a(n+1)=7/3(a(n+2)-7/4)=-1/3(a(n+1)-7/4)a3=5/3 a3-7/4=-1/12an-1是第三项是-1/12,公比-1/3的等...