Question

2006年北京市中学生数学竞赛高一年级决赛试题

递增数列{an}(n为下标)有a1=6，当n属于正整数数，n》=2时，an+a(n+1)=9/(an-a(n-1))+8,则a70=?

Answer 1

递增数列{an}，即an>a(n-1)>=a1=6 (n>=2)
当n>=2时,[an+a(n+1)][an-a(n-1)]=9+8[an-a(n-1)]
an^2-8an-[a(n-1)^2-8a(n-1)]-9=0
(an-4)^2=[a(n-1)-4]^2+9
设(an-4)^2=bn (n为正整数)
因为an>=6，所以an-4>=2>0，an-4=√bn,an=4+√bn
则n>=2时,bn=b(n-1)+9
b1=(a1-4)^2=4
{bn}为以4为首项,9为公差的等差数列
bn=9n-5(n为正整数)
所以an=4+√bn=4+√(9n-5)
a70=4+√(9*70-5)=4+25=29

Answer 2

5