ét ô ét , giúp em với ạ

 \begin{array}{l}
a,\ A\ =\ \dfrac{1}{3} +\dfrac{1}{6} +...+\dfrac{2}{n( n+1)} =\ \dfrac{1999}{2001}\\
=\ 2\left(\dfrac{1}{6} +\dfrac{1}{12} +..+\dfrac{1}{n( n+1)}\right) \ \\
=\ 2\left(\dfrac{1}{2.3} +\dfrac{1}{3.4} +\dfrac{1}{n( n+1)}\right) \ =\ 2.\left(\dfrac{1}{2} -\dfrac{1}{3} +\dfrac{1}{3} -\dfrac{1}{4} +..+\dfrac{1}{n} -\dfrac{1}{n+1}\right)\\
=\ 2\left(\dfrac{1}{2} -\dfrac{1}{n+1}\right) \ =\ 1\ -\dfrac{2}{n+1} \ =\ 1-\dfrac{2}{2001}\\
\Longrightarrow \ n+1\ =\ 2001\ \Longrightarrow \ n\ =\ 2000\\
b,\\
A=\ \dfrac{2n+5}{3n+1} \ \in \ N\ \Longrightarrow \ 3A\ =\ \dfrac{6n+15}{3n+1} \in N\Longrightarrow \ 2+\dfrac{13}{3n+1} \ \in N\\
\Longrightarrow \ 3n+1\ \in \ U( 13) \ =\ \{1,13\} \ ( \ do\ 3n+1\ \geqslant \ 1)\\
\Longrightarrow \ n\ \in \{0,\ 4\}\\
Thu\ lai\ \Longrightarrow \ n\ =\ 0\ \land \ n\ =\ 4
\end{array}