Ta có công thức: $\frac{1}{n}$-$\frac{1}{n+1}$

= $\frac{n+1}{n.( n+1)}$-$\frac{n}{n.( n+1)}$   

= $\frac{n+1-n}{n.( n+1)}$

= $\frac{1}{n.( n+1)}$  

⇒ $\frac{1}{n.( n+1)}$= $\frac{1}{n}$-$\frac{1}{n+1}$

Áp dụng vào, ta có: A= $\frac{1}{1.2}$+$\frac{1}{2.3}$+...+$\frac{1}{49.50}$

= 1-$\frac{1}{2}$+$\frac{1}{2}$-$\frac{1}{3}$+....+$\frac{1}{49}$-$\frac{1}{50}$

= 1-$\frac{1}{50}$

= $\frac{49}{50}$

Ta có $\frac{1}{1}$  - $\frac{1}{2}$ = $\frac{1.2-1}{1.2}$ = $\frac{1}{1.2}$

          $\frac{1}{2}$ - $\frac{1}{3}$ = $\frac{1.3-1.2}{2.3}$ = $\frac{1}{2.3}$

          $\frac{1}{3}$ - $\frac{1}{4}$ = $\frac{1.4-1.3}{3.4}$ = $\frac{1}{3.4}$

          $\frac{1}{49}$ - $\frac{1}{50}$ = $\frac{1.50-1.49}{49.50}$ = $\frac{1}{49.50}$

Mấy số khác cũng tương tự nên ta có

$\frac{1}{1.2}$ + $\frac{1}{2.3}$ + $\frac{1}{3.4}$ + ... + $\frac{1}{49.50}$

= $\frac{1}{1}$  - $\frac{1}{2}$ + $\frac{1}{2}$ - $\frac{1}{3}$ + $\frac{1}{3}$ - $\frac{1}{4}$ + ... + $\frac{1}{49}$ - $\frac{1}{50}$

= $\frac{1}{1}$ - $\frac{1}{50}$

= $\frac{1.50-1.1}{1.50}$

= $\frac{49}{50}$