Số xấu thế nhỉ?

\(u_n=v_n+\dfrac{\sqrt{5}-3}{2}\)

\(\Rightarrow v_{n+1}+\dfrac{\sqrt{5}-3}{2}=-\dfrac{1}{3+v_n+\dfrac{\sqrt{5}-3}{2}}\)

\(\Rightarrow\left\{{}\begin{matrix}v_1=u_1-\dfrac{\sqrt{5}-3}{2}=\dfrac{5-\sqrt{5}}{2}\\v_{n+1}=\dfrac{\dfrac{3-\sqrt{5}}{2}v_n}{\dfrac{3+\sqrt{5}}{2}+v_n}\end{matrix}\right.\)

\(v_n=\dfrac{1}{y_n}\Rightarrow\dfrac{1}{y_{n+1}}=\dfrac{\dfrac{3-\sqrt{5}}{2}.\dfrac{1}{y_n}}{\dfrac{3+\sqrt{5}}{2}+\dfrac{1}{y_n}}\)

\(\Rightarrow\dfrac{1}{y_{n+1}}=\dfrac{3-\sqrt{5}}{2y_n\left(\dfrac{3+\sqrt{5}}{2}+\dfrac{1}{y_n}\right)}=\dfrac{3-\sqrt{5}}{\left(3+\sqrt{5}\right)y_n+2}\)

\(\Leftrightarrow y_{n+1}=\dfrac{\left(3+\sqrt{5}\right)y_n}{3-\sqrt{5}}+\dfrac{2}{3-\sqrt{5}}\)

\(\Rightarrow\left\{{}\begin{matrix}y_1=\dfrac{1}{v_1}=\dfrac{2}{5-\sqrt{5}}\\y_{n+1}=\dfrac{3+\sqrt{5}}{3-\sqrt{5}}y_n+\dfrac{2}{3-\sqrt{5}}\end{matrix}\right.\)

\(z_n=y_n+\dfrac{\sqrt{5}}{5}\Rightarrow\left\{{}\begin{matrix}z_1=y_1+\dfrac{\sqrt{5}}{5}=\dfrac{5+3\sqrt{5}}{10}\\z_{n+1}=\dfrac{3+\sqrt{5}}{3-\sqrt{5}}z_n\end{matrix}\right.\)

\(\Rightarrow z_n:csn-co:\left\{{}\begin{matrix}z_1=\dfrac{5+3\sqrt{5}}{10}\\q=\dfrac{3+\sqrt{5}}{3-\sqrt{5}}\end{matrix}\right.\)

\(\Rightarrow z_{n+1}=\dfrac{5+3\sqrt{5}}{10}.\left(\dfrac{3+\sqrt{5}}{3-\sqrt{5}}\right)^n\)

\(\Rightarrow y_{n+1}=z_{n+1}-\dfrac{\sqrt{5}}{5}=\dfrac{5+3\sqrt{5}}{10}\left(\dfrac{3+\sqrt{5}}{3-\sqrt{5}}\right)^n-\dfrac{\sqrt{5}}{5}\)

\(v_{n+1}=\dfrac{1}{y_{n+1}}=\dfrac{1}{\dfrac{5+3\sqrt{5}}{10}\left(\dfrac{3+\sqrt{5}}{3-\sqrt{5}}\right)^n-\dfrac{\sqrt{5}}{5}}\)

\(u_{n+1}=v_{n+1}+\dfrac{\sqrt{5}-3}{2}=\dfrac{1}{\dfrac{5+3\sqrt{5}}{10}\left(\dfrac{3+\sqrt{5}}{3-\sqrt{5}}\right)^n-\dfrac{\sqrt{5}}{5}}+\dfrac{\sqrt{5}-3}{2}\)

Xét: 

\(u_{n+2}-u_{n+1}=\dfrac{1}{\dfrac{5+3\sqrt{5}}{10}.\left(\dfrac{3+\sqrt{5}}{3-\sqrt{5}}\right)^n.\left(\dfrac{3+\sqrt{5}}{3-\sqrt{5}}\right)-\dfrac{\sqrt{5}}{5}}+\dfrac{\sqrt{5}-2}{2}-\dfrac{1}{\dfrac{5+3\sqrt{5}}{10}\left(\dfrac{3+\sqrt{5}}{3-\sqrt{5}}\right)^n-\dfrac{\sqrt{5}}{5}}-\dfrac{\sqrt{5}-2}{2}\)

\(=\dfrac{1}{\dfrac{5+3\sqrt{5}}{10}\left(\dfrac{3+\sqrt{5}}{3-\sqrt{5}}\right)^n.\dfrac{3+\sqrt{5}}{3-\sqrt{5}}-\dfrac{\sqrt{5}}{5}}-\dfrac{1}{\dfrac{5+3\sqrt{5}}{10}\left(\dfrac{3+\sqrt{5}}{3-\sqrt{5}}\right)^n-\dfrac{\sqrt{5}}{5}}\)

\(=\dfrac{\dfrac{5+3\sqrt{5}}{10}\left(\dfrac{3+\sqrt{5}}{3-\sqrt{5}}\right)^n-\dfrac{5+3\sqrt{5}}{10}\left(\dfrac{3+\sqrt{5}}{3-\sqrt{5}}\right)^n.\left(\dfrac{3+\sqrt{5}}{3-\sqrt{5}}\right)}{.....}\)

\(=\dfrac{\dfrac{5+3\sqrt{5}}{10}\left(\dfrac{3+\sqrt{5}}{3-\sqrt{5}}\right)^n\left(1-\dfrac{3+\sqrt{5}}{3-\sqrt{5}}\right)}{....}=\dfrac{\dfrac{5+3\sqrt{5}}{10}\left(\dfrac{3+\sqrt{5}}{3-\sqrt{5}}\right)^n.\left(-\dfrac{5+3\sqrt{5}}{2}\right)}{...}< 0\)

\(\Rightarrow\) dãy giảm

\(\Rightarrow u_1>u_2>....>u_n\)

\(\Rightarrow\lim\limits u_n=1\)

Bn tham khảo đây nhé: https://diendantoanhoc.org/topic/140204-t%C3%A0i-li%E1%BB%87u-d%C3%A3y-s%E1%BB%91/

 Dễ thấy \(u_n>0,\forall n\inℕ^∗\). 

 Ta có \(u_{n+1}-u_n=\dfrac{u_n^2+2021}{2u_n}-u_n=\dfrac{2021-u_n^2}{2u_n}\)

 Với \(n\ge2\) thì \(u_n=\dfrac{u_{n-1}^2+2021}{2u_{n-1}}\) \(=\dfrac{u_{n-1}}{2}+\dfrac{2021}{2u_{n-1}}\) \(>2\sqrt{\dfrac{u_{n-1}}{2}.\dfrac{2021}{2u_{n-1}}}\) \(=\sqrt{2021}\)

Vậy \(u_n>\sqrt{2021},\forall n\ge2\), suy ra \(u_{n+1}-u_n=\dfrac{2021-u_n^2}{2u_n}< 0,\forall n\inℕ^∗\)

\(\Rightarrow\) Dãy \(\left(u_n\right)\) là dãy giảm. Mà \(u_n>\sqrt{2021}\)  \(\Rightarrow\left(u_n\right)\) có giới hạn hữu hạn. Đặt \(\lim\limits_{n\rightarrow+\infty}u_n=L\) \(\Rightarrow L=\dfrac{L^2+2021}{2L}\) \(\Leftrightarrow L=\sqrt{2021}\)

 Vậy \(\lim\limits_{n\rightarrow+\infty}u_n=\sqrt{2021}\)

Dễ thấy $u^n>0,\forall n\in N^{\ast }$. 

 Ta có $u^{n+1}-u^n=\genfrac{}{}{0ex}{}{u^n+2021}{2u^n}-u^n=\genfrac{}{}{0ex}{}{2021-u^n}{2u^n}$

 Với $n\ge 2$ thì $u^n=\genfrac{}{}{0ex}{}{u^{n-1}+2021}{2u^{n-1}}$ $=\genfrac{}{}{0ex}{}{u^{n-1}}{2}+\genfrac{}{}{0ex}{}{2021}{2u^{n-1}}$ $>2\sqrt{\genfrac{}{}{0ex}{}{u^{n-1}}{2}.\genfrac{}{}{0ex}{}{2021}{2u^{n-1}}}$ $=\sqrt{2021}$

Vậy $u^n>\sqrt{2021},\forall n\ge 2$, suy ra $u^{n+1}-u^n=\genfrac{}{}{0ex}{}{2021-u^n}{2u^n}<0,\forall n\in N^{\ast }$

$\Rightarrow$ Dãy $\left(u^n\right)$ là dãy giảm. Mà $u^n>\sqrt{2021}$  $\Rightarrow \left(u^n\right)$ có giới hạn hữu hạn. Đặt $\stackrel{\lim }{n\to +\infty }{u}^n=L$ $\Rightarrow L=\genfrac{}{}{0ex}{}{L^2+2021}{2L}$ $\iff L=\sqrt{2021}$

 Vậy $\stackrel{\lim }{n\to +\infty }{u}^n=\sqrt{2021}$
 

lim(n =) + ∞) u_n = \(\sqrt{2021}\) 

Ta sẽ chứng minh dãy bị chặn trên bởi 2

Thật vậy, với \(n=1;2\) thỏa mãn

Giả sử điều đó cũng đúng với \(n=k\) , tức \(u_k< 2\)

Ta cần chứng minh \(u_{k+1}< 2\)

Ta có: \(u_{k+1}=\sqrt{3u_k-2}< \sqrt{3.2-2}=2\) (đpcm)

Tương tự, ta cũng quy nạp được dễ dàng \(u_n>1\)

Mặt khác: \(u_n-u_{n-1}=\sqrt{3u_{n-1}-2}-u_{n-1}=\dfrac{3u_{n-1}-2-u_{n-1}^2}{\sqrt{3u_{n-1}-2}+u_{n-1}}\)

\(=\dfrac{\left(2-u_{n-1}\right)\left(u_{n-1}-1\right)}{\sqrt{3u_{n-1}-2}+u_{n-1}}>0\)

\(\Rightarrow u_n>u_{n-1}\Rightarrow\) dãy tăng

Dãy tăng và bị chặn trên nên có giới hạn hữu hạn.

Gọi giới hạn đó là k thì:

\(k=\sqrt{3k-2}\Leftrightarrow k=2\)

Lời giải:

Bằng quy nạp ta dễ chứng minh được $u_n< 1$

$u_{n+1}-u_n=\frac{1}{2-u_n}-u_n=\frac{(u_n-1)^2}{2-u_n}>0$ với mọi $u_n< 1$

$\Rightarrow u_{n+1}>u_n$. Vậy $(u_n)$ là dãy tăng và bị chặn trên. 

Gọi $\lim u_n=a$ thì $a=\frac{1}{2-a}\Rightarrow 2a-a^2=1$

$\Leftrightarrow (a-1)^2=0\Leftrightarrow a=1$

Đáp án B