Đáp án:
\[{u_n} = \frac{{{{6.5}^{n - 1}}}}{{{{3.5}^{n - 1}} - 1}}\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\left\{ \begin{array}{l}
{u_1} = 3\\
{u_{n + 1}} = \frac{{5{u_n}}}{{2{u_n} + 1}},\,\,\,\,\forall n
\end{array} \right.\\
{u_{n + 1}} = \frac{{5{u_n}}}{{2{u_n} + 1}}\\
\Leftrightarrow \frac{1}{{{u_{n + 1}}}} = \frac{{2{u_n} + 1}}{{5{u_n}}}\\
\Leftrightarrow \frac{1}{{{u_{n + 1}}}} = \frac{2}{5} + \frac{1}{{5{u_n}}}\\
\Leftrightarrow \frac{1}{{{u_{n + 1}}}} - \frac{1}{2} = \frac{1}{{5{u_n}}} - \frac{1}{{10}}\\
\Leftrightarrow \frac{1}{{{u_{n + 1}}}} - \frac{1}{2} = \frac{1}{5}\left( {\frac{1}{{{u_n}}} - \frac{1}{2}} \right)\\
{v_n} = \frac{1}{{{u_n}}} - \frac{1}{2} \Rightarrow \left\{ \begin{array}{l}
{v_1} = \frac{1}{{{u_1}}} - \frac{1}{2} = - \frac{1}{6}\\
{v_{n + 1}} = \frac{1}{5}{v_n}
\end{array} \right.\\
\Rightarrow {v_n} = {\left( {\frac{1}{5}} \right)^{n - 1}}{v_1} = {\left( {\frac{1}{5}} \right)^{n - 1}}.\left( { - \frac{1}{6}} \right) = - \frac{1}{{{{6.5}^{n - 1}}}}\\
\Rightarrow \frac{1}{{{u_n}}} - \frac{1}{2} = - \frac{1}{{{{6.5}^{n - 1}}}}\\
\Leftrightarrow \frac{1}{{{u_n}}} = \frac{1}{2} - \frac{1}{{{{6.5}^{n - 1}}}}\\
\Leftrightarrow \frac{1}{{{u_n}}} = \frac{{{{3.5}^{n - 1}} - 1}}{{{{6.5}^{n - 1}}}}\\
\Rightarrow {u_n} = \frac{{{{6.5}^{n - 1}}}}{{{{3.5}^{n - 1}} - 1}}
\end{array}\)
