Giải thích các bước giải:
B4:
a) Ta có:
$\begin{array}{l}
S = \dfrac{2}{{3.5}} + \dfrac{3}{{5.8}} + \dfrac{4}{{8.12}} + ... + \dfrac{{14}}{{93.107}} + \dfrac{{15}}{{107.122}} + \dfrac{{16}}{{122.138}}\\
= \dfrac{{5 - 3}}{{3.5}} + \dfrac{{8 - 5}}{{5.8}} + \dfrac{{12 - 8}}{{8.12}} + ... + \dfrac{{107 - 93}}{{93.107}} + \dfrac{{122 - 107}}{{107.122}} + \dfrac{{138 - 122}}{{122.138}}\\
= \left( {\dfrac{1}{3} - \dfrac{1}{5}} \right) + \left( {\dfrac{1}{5} - \dfrac{1}{8}} \right) + \left( {\dfrac{1}{8} - \dfrac{1}{{12}}} \right) + ... + \left( {\dfrac{1}{{93}} - \dfrac{1}{{107}}} \right) + \left( {\dfrac{1}{{107}} - \dfrac{1}{{122}}} \right) + \left( {\dfrac{1}{{122}} - \dfrac{1}{{138}}} \right)\\
= \dfrac{1}{3} - \dfrac{1}{{138}}\\
= \dfrac{{46 - 1}}{{138}}\\
= \dfrac{{45}}{{138}}\\
= \dfrac{{15}}{{46}}
\end{array}$
Vậy $S = \dfrac{{15}}{{46}}$
b) Ta có:
$\begin{array}{l}
A = \dfrac{n}{{n + 1}} + \dfrac{{n + 3}}{{n + 1}}\\
= \dfrac{{2n + 3}}{{n + 1}}\\
= \dfrac{{2\left( {n + 1} \right) + 1}}{{n + 1}}\\
= 2 + \dfrac{1}{{n + 1}}
\end{array}$
Để $A \in Z$
$\begin{array}{l}
\Leftrightarrow 2 + \dfrac{1}{{n + 1}} \in Z\\
\Leftrightarrow \dfrac{1}{{n + 1}} \in Z\\
\Leftrightarrow n + 1 \in U\left( 1 \right) = \left\{ { - 1;1} \right\}\left( {do:n + 1 \in Z} \right)\\
\Leftrightarrow n \in \left\{ { - 2;0} \right\}
\end{array}$
Vậy $n \in \left\{ { - 2;0} \right\}$ thỏa mãn đề.
