Đáp án:

Công thức  $\frac{4}{k(k+2)(k+4)}$ =$\frac{1}{k(k+2)}$ -$\frac{1}{(k+2)(k+4)}$ 

Ta có B=9($\frac{4}{1.3.5}$ +$\frac{4}{3.5.7}$ +...+$\frac{4}{25.27.29}$)

=9($\frac{1}{1.3}$- $\frac{1}{3.5}$+ $\frac{1}{3.5}$- $\frac{1}{5.7}$ +...+$\frac{1}{25.27}$ -$\frac{1}{27.29}$ )

=9($\frac{1}{3}$ -$\frac{1}{27.29}$)

=$\frac{260}{87}$ 

Giải thích các bước giải:

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

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

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

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

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

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

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

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