Question

Cho dãy u_n thỏa mản

\(\left\{{}\begin{matrix}u_1=1\\u_{n+1}=\dfrac{2u_n}{u_n+4}\end{matrix}\right.\) Với mọi n > bằng 1. Tìm CT SHTQ

Answer 1

Xét \(\dfrac{1}{u_{n+1}}=\dfrac{u_n+4}{2u_n}=\dfrac{1}{2}\left(1+\dfrac{4}{u_n}\right)\) (1)

Đặt \(\dfrac{1}{u_n}=x_n\)

(1) <=> \(x_{n+1}=\dfrac{1}{2}\left(4x_n+1\right)=2x_n+\dfrac{1}{2}\)

<=> \(x_{n+1}+\dfrac{1}{2}=2\left(x_n+\dfrac{1}{2}\right)\) (2) 

Đặt \(x_n+\dfrac{1}{2}=t_n\)

(2) <=> t_n+1 = 2.t_n => q = 2

Có: \(t_n=t_1.2^{n-1}\)

Mà \(t_1=x_1+\dfrac{1}{2}=\dfrac{1}{u_1}+\dfrac{1}{2}=\dfrac{3}{2}\)

=> \(t_n=\dfrac{3}{2}.2^{n-1}\)

=> \(x_n=\dfrac{3}{2}.2^{n-1}-\dfrac{1}{2}\)

=> \(u_n=\dfrac{2}{3.2^{n-1}-1}\)