Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\frac{1}{{{u_{n + 1}}}} = \frac{1}{{{u_n}}} + 2n + 2\\
\Rightarrow \frac{1}{{{u_{n + 1}}}} - \frac{1}{{{u_n}}} = 2n + 2\\
\Leftrightarrow \left\{ \begin{array}{l}
\frac{1}{{{u_2}}} - \frac{1}{{{u_1}}} = 2.1 + 2\\
\frac{1}{{{u_3}}} - \frac{1}{{{u_2}}} = 2.2 + 2\\
\frac{1}{{{u_4}}} - \frac{1}{{{u_3}}} = 2.3 + 2\\
......\\
\frac{1}{{{u_n}}} - \frac{1}{{{u_{n - 1}}}} = 2.\left( {n - 1} \right) + 2
\end{array} \right.\\
\Rightarrow \left( {\frac{1}{{{u_2}}} - \frac{1}{{{u_1}}}} \right) + \left( {\frac{1}{{{u_3}}} - \frac{1}{{{u_2}}}} \right) + \left( {\frac{1}{{{u_4}}} - \frac{1}{{{u_3}}}} \right) + .... + \left( {\frac{1}{{{u_n}}} - \frac{1}{{{u_{n - 1}}}}} \right) = \left( {2.1 + 2} \right) + \left( {2.2 + 2} \right) + \left( {2.3 + 2} \right) + .... + \left[ {2.\left( {n - 1} \right) + 2} \right]\\
\Leftrightarrow \frac{1}{{{u_n}}} - \frac{1}{{{u_1}}} = 2.\left( {1 + 2 + 3 + .... + n - 1} \right) + 2.\left( {n - 1} \right)\\
\Leftrightarrow \frac{1}{{{u_n}}} = \frac{1}{{{u_1}}} = 2.\left( {1 + 2 + 3 + ... + n - 1} \right) + 2.\left( {n - 1} \right)
\end{array}\)
