[tex] \displaystyle\\
\text{Folosim urmatoarea formula:}\\\\
\frac{3}{n(n+3)}=\frac{1}{n}-\frac{1}{n+3}\\\\
\text{Rezolvare:}\\\\
\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{3}{2012\cdot2015}=\\\\
=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+
\frac{1}{2012}-\frac{1}{2015}=\\\\
\text{Se reduc termenii asemenea.}\\\\
=\frac{1}{2}-\frac{1}{2015}=\frac{2015-2}{2\cdot 2015}=\boxed{\bf \frac{2013}{4030}}\\\\
\text{\bf cctd} [/tex]