4An+1-AnAn+1+2An=9
(4-An)An+1=9-2An
An+1=(9-2An)/(4-An)
=2+1/(4-An)
A2=7/3
A3=13/5
.
.
.
An=(6n-5)/(2n-1)
证明:
当n=1时
A1=(6*1-5)/(2*1-1)
=1
命题成立
假设当n=k时,命题成立
即 Ak=(6k-5)/(2k-1)
Ak+1=2+1/(4-Ak)
=2+1/[4-(6k-5)/(2k-1)]
=2+1/[(2k+1)/(2k-1)]
=2+(2k-1)/(2k+1)
=(6k+1)/(2k+1)
=[6(k+1)-5]/[2(k+1)-1]
所以命题成立
综上 An=(6n-5)/(2n-1) 成立