Lời giải:
$\frac{1}{4}< \frac{1}{1.2}$
$\frac{1}{9}< \frac{1}{2.3}$
$\frac{1}{16}< \frac{1}{3.4}$
....
$\frac{1}{2500}< \frac{1}{49.50}$
Cộng theo vế:
$A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{49}-\frac{1}{50}=1-\frac{1}{50}< 1$
Ta có đpcm.