\(S=\dfrac{1}{2018!\left(2019-2018\right)!}+\dfrac{1}{2016!\left(2019-2016\right)!}+...+\dfrac{1}{2!\left(2019-2\right)!}+\dfrac{1}{0!\left(2019-0!\right)}\)

\(\Rightarrow2019!.S=\dfrac{2019!}{2018!\left(2019-2018\right)!}+\dfrac{2019!}{2016!\left(2019-2016\right)!}+...+\dfrac{2019!}{2!\left(2019-2\right)!}+\dfrac{2019!}{0!\left(2019-0\right)!}\)

\(=C_{2019}^{2018}+C_{2019}^{2016}+...+C_{2019}^2+C_{2019}^0\)

\(=\dfrac{1}{2}\left(C_{2019}^0+C_{2019}^1+...+C_{2019}^{2018}+C_{2019}^{2019}\right)\)

\(=\dfrac{1}{2}.2^{2019}=2^{2018}\)

\(\Rightarrow S=\dfrac{2^{2018}}{2019!}\)

@Nguyễn Việt Lâm giúp em với

\(\dfrac{1}{u_n-1}=\dfrac{1}{\dfrac{2^n-5^n}{2^n+5^n}-1}=\dfrac{2^n+5^n}{-2.5^n}=-\dfrac{1}{2}\left[\left(\dfrac{2}{5}\right)^n+1\right]\)

\(\Rightarrow S_n=-\dfrac{1}{2}\left[\left(\dfrac{2}{5}\right)^1+\left(\dfrac{2}{5}\right)^2+...+\left(\dfrac{2}{5}\right)^n+n\right]\)

Lại có: \(\left(\dfrac{2}{5}\right)^1+\left(\dfrac{2}{5}\right)^2+...+\left(\dfrac{2}{5}\right)^n=\dfrac{2}{5}.\dfrac{1-\left(\dfrac{2}{5}\right)^n}{1-\dfrac{2}{5}}=\dfrac{2}{3}\left[1-\left(\dfrac{2}{5}\right)^n\right]\)

\(\Rightarrow S_n=-\dfrac{1}{2}\left[\dfrac{2}{3}-\dfrac{2}{3}\left(\dfrac{2}{5}\right)^n+n\right]=...\)

\(u_n-4u_{n-1}+3u_{n-2}=5.2^n\)

\(\Leftrightarrow u_n-u_{n-1}-3\left(u_{n-1}-u_{n-2}\right)=5.2^n\)

Đặt \(u_n-u_{n-1}=v_n\Rightarrow\left\{{}\begin{matrix}v_1=u_1-u_0=4\\v_n-3v_{n-1}=5.2^n\end{matrix}\right.\)

\(\Rightarrow v_n+10.2^n=3\left(v_{n-1}+10.2^{n-1}\right)\)

Đặt \(v_n+10.2^n=x_n\Rightarrow\left\{{}\begin{matrix}x_1=v_1+10.2^1=24\\x_n=3x_{n-1}\end{matrix}\right.\)

\(\Rightarrow x_n\) là CSN với công bội 3

\(\Rightarrow x_n=24.3^{n-1}\)

\(\Rightarrow v_n=x_n-10.2^n=24.3^{n-1}-10.2^n=8.3^n-10.2^n\)

\(\Rightarrow u_n-u_{n-1}=8.3^n-10.2^n\)

\(\Rightarrow u_n-12.3^n+20.2^n=u_{n-1}-12.3^{n-1}+20.2^{n-1}\)

Đặt \(u_n-12.3^n+20.2^n=y_n\Rightarrow\left\{{}\begin{matrix}y_1=u_1-12.3^1+20.2^1=7\\y_n=y_{n-1}=...=y_1=7\end{matrix}\right.\)

\(\Rightarrow u_n=12.3^n-20.2^n+7\)

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

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

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:

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

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

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

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

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

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

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

Xét số hạng tổng quát.

Cho k chạy từ 1 đến 2018 ta được:

Cho x=1 suy ra 

Chọn A.

Đặt \(\dfrac{u_n}{n+1}=v_n\)

\(GT\Rightarrow\left\{{}\begin{matrix}v_1=\dfrac{u_1}{1+1}=1\\v_{n+1}=\dfrac{1}{4}v_n,\forall n\in N\text{*}\end{matrix}\right.\)

\(\Rightarrow v_n=\dfrac{1}{4}^{n-1},\forall n\in N\text{*}\)

\(\Rightarrow u_n=\left(n+1\right).\dfrac{1}{4}^{n-1},\forall n\in N\text{*}\)