Cho dãy số xác định bởi

A

- Hướng dẫn giải

Đáp án A.

Ta có

$$\begin{gathered} \frac{n-1}{n^2+3n+2}=\frac{n-1}{\left(n+1\right)\left(n+2\right)}=\frac{A}{n+1}+\frac{B}{n+2} \\ \Rightarrow \left\{\begin{array}{l}A+B=1\\ 2A+B=-1\end{array}\right.\iff \left\{\begin{array}{l}A=-2\\ B=3\end{array}\right.. \end{gathered}$$ 

Lại có $3u_{n+1}=2u_n-\frac{2}{n+1}+\frac{3}{n+2}$

$\iff 3\left(u_{n+1}-\frac{1}{n+2}\right)=2\left(u_n-\frac{1}{n+1}\right).$

Đặt $v_n=u_n-\frac{1}{n+1}\Rightarrow v_1=\frac{1}{2}$

và $v_n=u_n-\frac{1}{n+1}\stackrel{}{\to }{v}_n$

là cấp số nhân với $v_1=\frac{1}{2};q=\frac{1}{3}$ 

$$\begin{gathered} \Rightarrow v_n=\frac{1}{2}.{\left(\frac{2}{3}\right)}^{n-1}=\frac{3}{4}.{\left(\frac{2}{3}\right)}^n \\ \to u_n=v_n+\frac{1}{n+1}=\frac{3}{4}.{\left(\frac{2}{3}\right)}^n+\frac{1}{n+1} \\ =\frac{2^{n-2}}{3^{n-1}}+\frac{1}{n+1}. \end{gathered}$$

 $$\begin{gathered} \Rightarrow u_{2018}=\left(\frac{2^{n-2}}{3^{n-1}}+\frac{1}{n+1}\right)\left|\begin{array}{l}\\ {}_{n=2018}\end{array}\right. \\ =\frac{2^{2016}}{3^{2017}}+\frac{1}{2019}. \end{gathered}$$