*Công thức tổng quát:

$\frac{1}{n(n+1)(n+2)}=$ $\frac{1}{2}.($ $\frac{1}{n(n+1)}-$ $\frac{1}{(n+1)(n+2)})$ 

Cống thức sử dụng trong bài:

$\frac{36}{1.3.5}=$ $\frac{18}{2}.($ $\frac{1}{1.3}-$ $\frac{1}{3.5})=9.($ $\frac{1}{1.3}-$ $\frac{1}{3.5})$ 

$B = \dfrac{36}{1.3.5} + \dfrac{36}{3.5.7} + \dfrac{36}{5.7.9} + ... + \dfrac{36}{25.27.29}$   

$\to B = 9.\left ( \dfrac{4}{1.3.5} + \dfrac{4}{3.5.7} + \dfrac{4}{5.7.9} + ... + \dfrac{4}{25.27.29} \right )$

$\to B = 9.\left [ \dfrac{1}{3}.\left ( \dfrac{4}{1.5} \right ) + \dfrac{1}{5}.\left ( \dfrac{4}{3.7} \right ) + \dfrac{1}{7}.\left ( \dfrac{4}{5.9} \right ) + ... + \dfrac{1}{27}.\left ( \dfrac{4}{25.29} \right ) \right ]$

$\to B = 9.\left [ \dfrac{1}{3}.\left ( 1 - \dfrac{1}{5} \right ) + \dfrac{1}{5}.\left ( \dfrac{1}{3} - \dfrac{1}{7} \right ) + \dfrac{1}{7}.\left ( \dfrac{1}{5} - \dfrac{1}{9} \right ) + ... + \dfrac{1}{27}.\left ( \dfrac{1}{25} - \dfrac{1}{29} \right ) \right ]$

$\to B = 9.\left ( \dfrac{1}{1.3} - \dfrac{1}{3.5} + \dfrac{1}{3.5} - \dfrac{1}{5.7} + \dfrac{1}{5.7} - \dfrac{1}{7.9} + ... + \dfrac{1}{25.27} - \dfrac{1}{27.29} \right )$

$\to B = 9.\left ( \dfrac{1}{1.3}  - \dfrac{1}{27.29} \right )$

$\to B = 3 - \dfrac{9}{783}$

$\to B = \dfrac{260}{87}$