- Hướng dẫn giải

Ta có:

$$\begin{gathered} \begin{array}{l}3\left(n+2\right)x_{n+1}^2\\ =2\left(n+1\right)x_{n+1}^2+\left(n+4\right)\end{array} \\ \forall n\ge 21 \end{gathered}$$

$$\begin{gathered} \begin{array}{l}\iff 3\left(n+2\right)x_{n+1}^2\\ =2\left(n+1\right)x_{n+1}^2-2\left(n+1\right)\\ +3\left(n-2\right),\forall n\ge 21\end{array} \\ \iff 3\left(n+2\right)\left(x_{n+1}^2-1\right) \\ =2\left(n+1\right)\left(x_n^2-1\right) \end{gathered}$$

Đặt $y_n=x_n^2-1$. Khi đó

$y_{n+1}=\frac{2}{3}.\frac{n+1}{n+2}{y}_n$

Suy ra

$$\begin{gathered} y_{n+1}=\frac{2\left(n+1\right)}{3\left(n+2\right)}.\frac{2n}{3\left(n+1\right)} \\ ={\left(\frac{2}{3}\right)}^{n+1}.\frac{1}{n+2}{y}_1 \end{gathered}$$

hay $lim{y}_n=0$. Vậy $lim{x}_n$ = 1

Đáp án cần chọn là B