Cho dãy số

D

- Hướng dẫn giải

Ta có $u_2=\frac{1}{3}$

Với $n\ge 3$ ta có

$$\begin{gathered} \left[u_1+2u_2+..+\left(n-1\right)u_{n-1}\right] \\ +n{u}_n=n\left(n^2-1\right)u_n+n{u}_n \\ =n^3{u}_n \\ \Rightarrow n{u_n}^3=n{u}_n+{\left(n-1\right)}^3{u}_{n-1} \\ \Rightarrow \frac{u_n}{u_{n-1}}=\frac{{\left(n-1\right)}^3}{n^3-n} \\ ={\left(\frac{n-1}{n}\right)}^2\left(\frac{n}{n+1}\right)\left(1\right) \end{gathered}$$

Từ (1) suy ra

$$\begin{gathered} \frac{u_n}{u_2}=\frac{u_n}{u_{n-1}}.\frac{u_{n-1}}{u_{n-2}}...\frac{u_3}{u_2} \\ =\left[{\left(\frac{n-1}{n}\right)}^2.{\left(\frac{n-1}{n-2}\right)}^2..{\left(\frac{2}{3}\right)}^2\right] \\ \left(\frac{n}{n-1}.\frac{n-1}{n}...\frac{3}{4}\right) \\ =\frac{12}{n^2\left(n+1\right)} \\ \Rightarrow u_n=\frac{4}{n^2\left(n+1\right)} \end{gathered}$$

Vậy $lim{\left(n+2018\right)}^3{U}_n$= 4

Đáp án D