[tex]\frac{1}{n(n+1)}=\frac{n+1-n}{n(n+1)}=\frac{n+1}{n(n+1)}-\frac{n}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}\\
S=\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+...+\frac{1}{49\cdot 50}\\
S=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\\
Termenii\ se\ reduc\ si\ mai\ ramane:\\
S=1-\frac{1}{50}\\
\boxed{S=\frac{49}{50}}[/tex]
