$\dfrac{3}{1× 3} + \dfrac{3}{3×5} + \dfrac{3}{5×7} + ... + \dfrac{3}{47×49} + \dfrac{3}{49×51}$
$=\dfrac{3}{2}\bigg(\dfrac{2}{1× 3} + \dfrac{2}{3×5} + \dfrac{2}{5×7} + ... + \dfrac{2}{47×49} + \dfrac{2}{49×51}\bigg)$
$=\dfrac{3}{2}\bigg(\dfrac{1}{1} - \dfrac{1}{3} + \dfrac{1}{3} - \dfrac{1}{5} + \dfrac{1}{5} ... + \dfrac{1}{49} - \dfrac{1}{51}\bigg)$
$=\dfrac{3}{2}\bigg(1 - \dfrac{1}{51}\bigg)$
$=\dfrac{3}{2} × \dfrac{50}{51}$
$=\dfrac{25}{17}$
=> Chọn ý C
