CT $\frac{4}{k(k+2)(k+4)}$ =$\frac{1}{k(k+2)}$ -$\frac{1}{(k+2)(k+4)}$
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($\frac{1}{3}$ -$\frac{1}{27.29}$)
=$\frac{260}{87}$
Kl