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[Coppy dera]

Cho $a_{n+1}=\frac{a_{n}}{n.a_{n}+1}$

Tính $a_{64}$

[Dung Du Duong]

Cho $a_{n+1}=\frac{a_{n}}{n.a_{n}+1}$

Tính $a_{64}$

$a_{n+1}=\frac{a_{n}}{n.a_{n}+1}$ <==> $\frac{1}{a_{n+1}}=n+\frac{1}{a_n}$

Đặt $\frac{1}{a_n} = u_n$ ==> $u_{n+1}=n+u_n=n+n-1+u_{n-1}=...=n+n-1+n-2+...+1+u_1=\frac{n(n+1)}{2}+u_1$ ==> $u_n=\frac{n(n-1)}{2}+u_1$

Lưu ý: đề bài cần cho thêm $a_1$  thì mới làm đc   

Edited by Dung Du Duong, 15-03-2016 - 19:16.