[tex]\displaystyle\\
\text{folosim urmatoarea formula:}\\\\
\frac{1}{n(n+1)}= \frac{1}{n}- \frac{1}{n+1}\\\\
\frac{1}{12\times 13} +\frac{1}{13\times 14} +\frac{1}{14\times 15}+\cdots +\frac{1}{98\times 99}+\frac{1}{99\times 100}=\\\\
=\frac{1}{12}-\frac{1}{13}+ \frac{1}{13}-\frac{1}{14}+\frac{1}{14}-\frac{1}{15}+\cdots +
\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}=\\\\
=\frac{1}{12}-\frac{1}{100}=\frac{25}{12\cdot 25}-\frac{3}{100\cdot 3}=
\frac{25+3}{300}=\frac{28}{300}=\boxed{\bf \frac{7}{75}}[/tex]
