Đặt \(u_n=v_n+1\Rightarrow v_{n+1}+1=\dfrac{2017+v_n+1}{2019-\left(v_n+1\right)}=\dfrac{2018+v_n}{2018-v_n}\)

\(\Rightarrow v_{n+1}=\dfrac{2018+v_n}{2018-v_n}-1=\dfrac{2v_n}{2018-v_n}\Rightarrow\dfrac{1}{v_{n+1}}=1009\dfrac{1}{v_n}-\dfrac{1}{2}\)

Đặt \(\dfrac{1}{v_n}=x_n\Rightarrow\left\{{}\begin{matrix}x_1=\dfrac{1}{v_1}=\dfrac{1}{u_1-1}=1\\x_{n+1}=1009x_n-\dfrac{1}{2}\end{matrix}\right.\)

\(\Rightarrow x_{n+1}-\dfrac{1}{2016}=1009\left(x_n-\dfrac{1}{2016}\right)\)

\(\Rightarrow x_n-\dfrac{1}{2016}\) là CSN với công bội 1009 \(\Rightarrow x_n-\dfrac{1}{2016}=\dfrac{2015}{2016}.1009^{n-1}\)

\(\Rightarrow x_n=\dfrac{2015}{2016}1009^{n-1}+\dfrac{1}{2016}\) 

\(\Rightarrow u_n=v_n+1=\dfrac{1}{x_n}+1=\dfrac{2016}{2015.1009^{n-1}+1}+1\)

\(\Rightarrow\lim\left(u_n\right)=1\)

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

\(\Rightarrow\left(n+1\right)u_{n+1}=nu_n+n^2+2n+1\)

\(\Rightarrow\left(n+1\right)u_{n+1}-\dfrac{1}{3}\left(n+1\right)^3-\dfrac{1}{2}\left(n+1\right)^2-\dfrac{1}{6}\left(n+1\right)=n.u_n-\dfrac{1}{3}n^3-\dfrac{1}{2}n^2-\dfrac{1}{6}n\)

Đặt \(v_n=u.u_n-\dfrac{1}{3}n^3-\dfrac{1}{2}n^2-\dfrac{1}{6}n\Rightarrow\left\{{}\begin{matrix}v_1=1-\dfrac{1}{3}-\dfrac{1}{2}-\dfrac{1}{6}=0\\v_{n+1}=v_n=...=v_1=0\end{matrix}\right.\)

\(\Rightarrow n.u_n-\dfrac{1}{3}n^3-\dfrac{1}{2}n^2-\dfrac{1}{6}n=0\)

\(\Rightarrow u_n=\dfrac{1}{3}n^2+\dfrac{1}{2}n+\dfrac{1}{6}=\dfrac{\left(n+1\right)\left(2n+1\right)}{6}\)

\(u_{n+1}=\sqrt{1+u_n^2}\left(1\right)\)

\(u_1=3=\sqrt{9}\)

\(u_2=\sqrt{1+u_1^2}=\sqrt{10}\)

\(u_3=\sqrt{1+u_2^2}=\sqrt{11}\)

...

Dự đoán công thức:\(u_n=\sqrt{n+8}\),\(n\ge1\) (*)

Thật vậy 

+)\(n=1,(*)\)\(\Leftrightarrow u_1=3\) (lđ)

+)Giả sử (*) đúng với mọi \(n=k,k>1\)

\((*)\Leftrightarrow u_k=\sqrt{k+8}\)

+)\(n=k+1,\) thay vào (1) có: \(u_{k+2}=\sqrt{1+u^2_{k+1}}=\sqrt{1+\left(\sqrt{1+u_k^2}\right)^2}=\sqrt{2+u^2_k}=\sqrt{2+k+8}=\sqrt{10+k}\)

\(\Rightarrow\)(*) đúng với n=k+1

Vậy CTSHTQ: \(u_n=\sqrt{n+8}\), \(n\ge1\)

\(u_{n+1}=\dfrac{2u_n}{u_n+4}\Leftrightarrow\dfrac{1}{u_{n+1}}=\dfrac{1}{2}+\dfrac{2}{u_n}\)

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 \(\Rightarrow x_n=\dfrac{3}{2}.2^{n-1}=3.2^{n-2}\)

\(\Leftrightarrow v_n=x_n-\dfrac{1}{2}=3.2^{n-2}-\dfrac{1}{2}\)

\(\Rightarrow u_n=\dfrac{1}{v_n}=\dfrac{1}{3.2^{n-2}-\dfrac{1}{2}}=\dfrac{2}{3.2^{n-1}-1}\)

Cho \(\left\{{}\begin{matrix}u_1=1\\u_{n+1}=2u_n+6\end{matrix}\right.\)

Tìm số hạng tổng quát của dãy số sau

Đặt \(\dfrac{u_n}{n+1}=v_n\)

\(GT\Rightarrow\left\{{}\begin{matrix}v_1=\dfrac{u_1}{1+1}=1\\v_{n+1}=\dfrac{1}{4}v_n,\forall n\in N\text{*}\end{matrix}\right.\)

\(\Rightarrow v_n=\dfrac{1}{4}^{n-1},\forall n\in N\text{*}\)

\(\Rightarrow u_n=\left(n+1\right).\dfrac{1}{4}^{n-1},\forall n\in N\text{*}\)

\(u_{n+1}=\dfrac{3}{2}\left(u_n-\dfrac{n+4}{\left(n+1\right)\left(n+2\right)}\right)=\dfrac{3}{2}\left(u_n-\dfrac{3}{n+1}+\dfrac{2}{n+2}\right)\)

\(\Leftrightarrow u_{n+1}-\dfrac{3}{n+1+1}=\dfrac{3}{2}\left(u_n-\dfrac{3}{n+1}\right)\)

Đặt \(u_n-\dfrac{3}{n+1}=v_n\Rightarrow\left\{{}\begin{matrix}v_1=u_1-\dfrac{3}{2}=-\dfrac{1}{2}\\v_{n+1}=\dfrac{3}{2}v_n\end{matrix}\right.\)

\(\Rightarrow v_n\) là CSN với công bội \(\dfrac{3}{2}\)

\(\Rightarrow v_n=-\dfrac{1}{2}\left(\dfrac{3}{2}\right)^{n-1}\)

\(\Rightarrow u_n=-\dfrac{1}{2}\left(\dfrac{3}{2}\right)^{n-1}+\dfrac{3}{n+1}\)