Bài $B = \dfrac{1}{1.2.3} + \dfrac{1}{2.3.4} + \dfrac{1}{3.4.5} + ... + \dfrac{1}{18.19.20}$ lại phải lấy $2.B$ vì để ý ở dưới mẫu của các phân số lần lượt có $3 - 1 = 2$, $4 - 2 = 2$, $5 - 3 = 2$,...
$B = \dfrac{36}{1.3.5} + \dfrac{36}{3.5.7} + \dfrac{36}{5.7.9} + ... + \dfrac{36}{25.27.29}$
$\to B = 9.\left ( \dfrac{4}{1.3.5} + \dfrac{4}{3.5.7} + \dfrac{4}{5.7.9} + ... + \dfrac{4}{25.27.29} \right )$
$\to B = 9.\left [ \dfrac{1}{3}.\left ( \dfrac{4}{1.5} \right ) + \dfrac{1}{5}.\left ( \dfrac{4}{3.7} \right ) + \dfrac{1}{7}.\left ( \dfrac{4}{5.9} \right ) + ... + \dfrac{1}{27}.\left ( \dfrac{4}{25.29} \right ) \right ]$
$\to B = 9.\left [ \dfrac{1}{3}.\left ( 1 - \dfrac{1}{5} \right ) + \dfrac{1}{5}.\left ( \dfrac{1}{3} - \dfrac{1}{7} \right ) + \dfrac{1}{7}.\left ( \dfrac{1}{5} - \dfrac{1}{9} \right ) + ... + \dfrac{1}{27}.\left ( \dfrac{1}{25} - \dfrac{1}{29} \right ) \right ]$
$\to B = 9.\left ( \dfrac{1}{1.3} - \dfrac{1}{3.5} + \dfrac{1}{3.5} - \dfrac{1}{5.7} + \dfrac{1}{5.7} - \dfrac{1}{7.9} + ... + \dfrac{1}{25.27} - \dfrac{1}{27.29} \right )$
$\to B = 9.\left ( \dfrac{1}{1.3} - \dfrac{1}{27.29} \right )$
$\to B = 3 - \dfrac{9}{783}$
$\to B = \dfrac{260}{87}$
