在数列an中,已知a1=1,a(n+2)=1/(an+1),a100=a96,则a9+a10=?

Question

在数列an中,已知a1=1,a(n+2)=1/(an+1),a100=a96,则a9+a10=?

Answer 1

a9 = 1/(a7 +1)
= 1/[ 1/(a5+1) +1 ]
= (a5+1)/(a5+2)
= [ 1/(a3+1) +1 ]/[ 1/(a3+1) + 2]
= (a3+2)/(a3+3)
= (a1+3)/(a1+4)
= 4/5
a100= 1/(a98+1)
= 1/[1/(a96+1) + 1]
= (a100+1)/(a100+2)
(a100)^2+a100-1=0
a100=(-1+ √5)/2
a12 =a100= (-1+ √5)/2
a12 = 1/(a10 +1)
(-1+ √5)/2 =1/(a10 +1)
(-1+ √5)(a10 +1) = 2
a10 = (3-√5)/(-1+√5)
= (3-√5).(1+√5)/ 4
= (-1+√5)/2
a9+a10 = 4/5 + (-1+√5)/2