Lời giải:
Đặt $A=\dfrac{1}{3×5}+\dfrac{1}{5×7}+....+\dfrac{1}{99×101}$
$⇒A=\dfrac{1}{2}(\dfrac{2}{3×5}+\dfrac{2}{5×7}+....+\dfrac{2}{99×101})$
Áp dụng công thức:$\dfrac{k}{n(n+k)}=\dfrac{1}{n}-\dfrac{1}{n+k}$
$⇒A=\dfrac{1}{2}(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+....+\dfrac{1}{99}-\dfrac{1}{101}) $
$⇒A=\dfrac{1}{2}(\dfrac{1}{3}-\dfrac{1}{101})$
$⇒A=\dfrac{1}{6}-\dfrac{1}{202}$
$⇒A=\dfrac{101-3}{606}=\dfrac{98}{606}=\dfrac{49}{303}$
