Question

已知各项均为正数的数列{an}满足:a1=a3,a2=1,an+2=1/1+an,则a9+a10=？？

Answer 1

a(3)=a(1+2)=1/[1+a(1)]=a(1),
1=a(1)+[a(1)]^2,
0 = [a(1)]^2 + a(1) - 1,
Delta = 1 + 4 = 5. a(1)=[-1+5^(1/2)]/2, 或a(1)=[-1-5^(1/2)]/2<0(舍去）。
a(1)=[5^(1/2)-1]/2=a(3).

a(n+2)=1/[1+a(n)],
a(2n-1+2)=a(2n+1)=1/[1+a(2n-1)],
a(2n+2)=1/[1+a(2n)],

a(2n+1)=1/[1+a(2n-1)], a(1)=[5^(1/2)-1]/2=a(3),
由归纳法可证，a(2n-1)=a(1)=[5^(1/2)-1]/2.
a(9)=[5^(1/2)-1]/2.

a(2n+2)=1/[1+a(2n)], a(2)=1,
a(4)=1/[1+1]=1/2.
a(6)=1/[1+1/2]=2/3.
a(8)=1/[1+2/3]=3/5.
a(10)=1/[1+3/5]=5/8.

a(9)+a(10)=[5^(1/2)-1]/2 + 5/8 = (1/2)5^(1/2) + 1/8

Answer 2

你这个题目确定是这样的？首先a2就不满足an+2=1/1+an。

追问

我们的期末考题目