[tex]Forma\ generala:\frac{1}{n(n+1)}=\frac{n+1-n}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}\\
Revenind:\\
\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{n(n+1)}=\frac{1003}{1004}\\
\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}+...+\frac{1}{n(n+1)}=\frac{1003}{1004}\\
1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{n}-\frac{1}{n+1}=\frac{1003}{1004}\\
1-\frac{1}{n+1}=\frac{1003}{1004}\\
\frac{n}{n+1}=\frac{1003}{1004}\Rightarrow \bold{n=1003}[/tex]
