Chọn A

Phương pháp: Tìm công thức số hạng tổng quát

Cách giải: Ta có:

u ( 1 ) = 1

u ( 2 ) = u ( 1 ) + u ( 1 ) = 2 u ( 1 ) + 1

u ( 3 ) = u ( 2 ) + u ( 1 ) = 3 u ( 1 ) + 1 + 2

u ( 4 ) = u ( 3 ) + u ( 1 ) = 4 u ( 1 ) + 1 + 2 + 3

. . .

u ( 2017 ) = u ( 2016 ) + u ( 1 ) = 2017 u ( 1 ) + 1 + 2 + 3 . . . + 2016

⇒ u ( 2017 ) = 1 + 2 + 3 . . . + 2016 + 2017 = 2035153

\(\begin{cases} u_1 = 4 \\ u_{n+1} = \frac{3n}{n+1} u_n - \frac{2n^2 + 6n + 3}{n^2 (n+1)^3} \end{cases}\)

=>\(\frac{(n+1) u_{n+1}}{3^{n+1}} = \frac{n u_n}{3^n} - \frac{2n^2 + 6n + 3}{3^{n+1} \cdot n^2 (n+1)^2}\)

Đặt \(v_n = \frac{n u_n}{3^n}\)

=>\(\) \(v_1 = \frac{1 \cdot 4}{3^1} = \frac{4}{3}\)

Hệ thức sẽ trở thành:

\(v_{n+1} - v_n = -\frac{2n^2 + 6n + 3}{3^{n+1} \cdot n^2 (n+1)^2}\)

\(\frac{2n^2 + 6n + 3}{3^{n+1} n^2 (n+1)^2} = \frac{3(n+1)^2 - n^2}{3^{n+1} n^2 (n+1)^2} = \frac{1}{3^n \cdot n^2} - \frac{1}{3^{n+1} (n+1)^2}\)

=>\(v_{n+1} - v_n = \frac{1}{3^{n+1} (n+1)^2} - \frac{1}{3^n \cdot n^2}\)

Cộng các sai phân tử từ 1 đến n-1, ta có:

\(\sum_{k=1}^{n-1} (v_{k+1} - v_k) = \sum_{k=1}^{n-1} \left( \frac{1}{3^{k+1} (k+1)^2} - \frac{1}{3^k \cdot k^2} \right)\)

=>\(v_n - v_1 = \frac{1}{3^n \cdot n^2} - \frac{1}{3^1 \cdot 1^2}\)

\(v_n - \frac{4}{3} = \frac{1}{3^n \cdot n^2} - \frac{1}{3} \implies v_n = 1 + \frac{1}{3^n \cdot n^2}\)

=>\(\frac{n u_n}{3^n}=1+\frac{1}{3^n \cdot n^2}\)

=>\(u_{n}=\frac{3^n}{n}+\frac{1}{n^3}\)

\(\frac{n u_n}{4} = \frac{3^n \cdot v_n}{4} = \frac{3^n}{4} \left( 1 + \frac{1}{3^n \cdot n^2} \right) = \frac{3^n}{4} + \frac{1}{4n^2}\)

=>\(\lim \left( \frac{n u_n}{4} \right) = +\infty\)

Đề không cho sẵn dãy tăng à? Vậy phải chứng minh nó tăng trước

\(u_{n+1}=\dfrac{u_n^2+2018u_n+1}{2020}\)

\(u_{n+1}-u_n=\dfrac{u_n^2+2018u_n+1}{2020}-u_n=\dfrac{\left(u_n-1\right)^2}{2020}\ge0\) \(\Rightarrow\) dãy tăng và không bị chặn trên \(\Rightarrow lim\left(u_n\right)=+\infty\)

\(\Rightarrow2020u_{n+1}=u_n^2+2018u_n+1\)

\(\Leftrightarrow2020u_{n+1}-2020=u_n^2+2018u_n-2019\)

\(\Leftrightarrow2020\left(u_{n+1}-1\right)=\left(u_n+2019\right)\left(u_n-1\right)\)

\(\Rightarrow\dfrac{1}{2020\left(u_{n+1}-1\right)}=\dfrac{1}{\left(u_n+2019\right)\left(u_n-1\right)}=\dfrac{1}{2020}\left(\dfrac{1}{u_n-1}-\dfrac{1}{u_n+2019}\right)\)

\(\Rightarrow\dfrac{1}{u_n+2019}=\dfrac{1}{u_n-1}-\dfrac{1}{u_{n+1}-1}\)

Thế n=1;2;...;n ta được:

\(\dfrac{1}{u_1+2019}=\dfrac{1}{u_1-1}-\dfrac{1}{u_2-1}\)

\(\dfrac{1}{u_2+2019}=\dfrac{1}{u_2-1}-\dfrac{1}{u_3-1}\)

...

\(\dfrac{1}{u_n+2019}=\dfrac{1}{u_n-1}-\dfrac{1}{u_{n+1}-1}\)

Cộng vế: \(S_n=\dfrac{1}{u_n-1}-\dfrac{1}{u_{n+1}-1}=\dfrac{1}{2018}-\dfrac{1}{u_{n+1}-1}\)

\(\Rightarrow\lim\left(S_n\right)=\dfrac{1}{2018}-\dfrac{1}{\infty}=\dfrac{1}{2018}\)