Cho dãy số xác định bởi $\Large u_1=1$, $\Large u_{n+1}=\dfrac{1}{3}\l

Cho dãy số xác định bởi $\Large u_1=1$, $\Large u_{n+1}=\dfrac{1}{3}\left(2u_n+\dfrac{n-1}{n^2+3n+2}\right)$, $\Large n\in \mathbb{N^*}$. Khi đó $\Large u_{2018}$ bằng

• A.

  $\Large \dfrac{2^{2016}}{3^{2017}}+\dfrac{1}{2019}$.

• B.

  $\Large \dfrac{2^{2018}}{3^{2017}}+\dfrac{1}{2019}$.

• C.

  $\Large \dfrac{2^{2017}}{3^{2016}}+\dfrac{1}{2019}$.

• D.

  $\Large \dfrac{2^{2017}}{3^{2018}}+\dfrac{1}{2019}$.

Ta có $\Large u_{n+1}=\dfrac{1}{3}\left(2u_1+\dfrac{n-1}{n^2+3n+2}\right)$ $\Large \Leftrightarrow u_{n+1}=\dfrac{n-1}{3(n^2+3n+2)}+\dfrac{2}{3}u_n$

$\Large \Leftrightarrow u_{n+1}=\dfrac{2}{3}u_n+\dfrac{1}{3}\left[\dfrac{3}{n+2}-\dfrac{2}{n-1}\right]$

$\Large \Leftrightarrow u_{n+1}-\dfrac{1}{n+2}=\dfrac{2}{3}u_n-\dfrac{2}{3(n+1)}=\dfrac{2}{3}\left[u_1+\dfrac{1}{n+1}\right]$.

Đặt $\Large v_{n+1}=u_{n+1}-\dfrac{1}{n+2}$ $\Large \Rightarrow v_{n+1}=\dfrac{2}{3}v_n$ nên $\Large (v_n)$ là cấp số nhân với $\Large v_1=u_1-\dfrac{1}{1+1}=\dfrac{1}{2}$, công bội là $\Large q=\dfrac{2}{3}$ $\Large \Rightarrow v_n=v_1.\left(\dfrac{2}{3}\right)^{n-1}$ $\Large =\dfrac{2^{n-2}}{3^{n-1}}=u_n-\dfrac{1}{n+1}$

$\Large \Rightarrow u_n=\dfrac{2^{n-2}}{3^{n-1}}+\dfrac{1}{n+1}$.

Vậy $\Large u_{2018}=\dfrac{2^{2016}}{3^{2017}}+\dfrac{1}{2019}$.

Chọn đáp án A.