Cho dãy số xác định bởi u1 = 2018 và u( n + 1) = căn (un^2 + n^2 + 2018) > và bằng

* Hướng dẫn giải

Đáp án C

Ta có: $\left\{\begin{array}{l}{u}_1=2018\\ {u^2}_{n+1}={u^2}_n+n^2+2018,n\ge 1\end{array}\right.$

Đặt $v_n={u^2}_n\Rightarrow \left\{\begin{array}{l}{v}_1={2018}^2\\ v_{n+1}=v_n+n^2+2018\end{array}\right.$

Ta có :$\left\{\begin{array}{l}{v}_1={2018}^2\\ v_2=v_1+1^2+2018\\ v_3=v_2+2^2+2018\\ \mathrm{..........}\\ v_n=v_{n-1}+{\left(n-1\right)}^2+2018\end{array}\right.$

$$\begin{gathered} \Rightarrow v_n={2018}^2+2018\left(n-1\right)+\left(1+2+\mathrm{...}+n-1\right) \\ +\left(1^2+2^2+\mathrm{...}+{\left(n-1\right)}^2\right) \end{gathered}$$

Trong đó ta có:

$1^2+2^2+\mathrm{...}+{\left(n-1\right)}^2=\frac{\left(n-1\right)n\left(2n-1\right)}{6}$

Do đó

$$\begin{gathered} v_n={2018}^2+2018\left(n-1\right)+\frac{\left(n-1\right)n\left(2n-1\right)}{6} \\ \Rightarrow v_{21}=4115554\Rightarrow u_{21}=\sqrt{v_{21}}\approx 2028. \end{gathered}$$