Cho dãy số \((u_n)\) có \(\left\{ \begin{array}{l} {u_1} = 1\\ {u_{n + 1}} = 3n.{u_n} \end{array} \right.,\forall n \in {N^*}.\) Tính \({u_{^{2019}}}\)

* Hướng dẫn giải

\(\begin{array}{l}
\frac{{{u_{n + 1}}}}{{{u_n}}} = 3n\\
\frac{{{u_2}}}{{{u_1}}} = 3.1\\
\frac{{{u_3}}}{{{u_2}}} = 3.2\\
\frac{{{u_4}}}{{{u_3}}} = 3.3\\
..............\\
\frac{{{u_{2019}}}}{{{u_{2018}}}} = 3.2018 \Rightarrow \frac{{{u_{2019}}}}{{{u_1}}} = {3^{2018}}.2018! \Rightarrow {u_{2019}} = {3^{2018}}.2018!.
\end{array}\)