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\(u_{n+1}=\dfrac{2u_n}{u_n+4}\Leftrightarrow\dfrac{1}{u_{n+1}}=\dfrac{1}{2}+\dfrac{2}{u_n}\)
Đặt \(v_n=\dfrac{1}{u_n}\Rightarrow\left\{{}\begin{matrix}v_1=1\\v_{n+1}=2v_n+\dfrac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}v_1=1\\v_{n+1}+\dfrac{1}{2}=2\left(v_n+\dfrac{1}{2}\right)\end{matrix}\right.\)
Đặt \(v_n+\dfrac{1}{2}=x_n\Rightarrow\left\{{}\begin{matrix}x_1=\dfrac{3}{2}\\x_{n+1}=2x_n\end{matrix}\right.\)
\(\Rightarrow x_n\) là CSN với công bội 2 \(\Rightarrow x_n=\dfrac{3}{2}.2^{n-1}=3.2^{n-2}\)
\(\Leftrightarrow v_n=x_n-\dfrac{1}{2}=3.2^{n-2}-\dfrac{1}{2}\)
\(\Rightarrow u_n=\dfrac{1}{v_n}=\dfrac{1}{3.2^{n-2}-\dfrac{1}{2}}=\dfrac{2}{3.2^{n-1}-1}\)
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) <=> tn+1 = 2.tn => q = 2
Có: \(t_n=t_1.2^{n-1}\)
Mà \(t_1=x_1+\dfrac{1}{2}=\dfrac{1}{u_1}+\dfrac{1}{2}=\dfrac{3}{2}\)
=> \(t_n=\dfrac{3}{2}.2^{n-1}\)
=> \(x_n=\dfrac{3}{2}.2^{n-1}-\dfrac{1}{2}\)
=> \(u_n=\dfrac{2}{3.2^{n-1}-1}\)
