The value of $$\frac{1}{1^2.3^2} + \frac{2}{3^2.5^2} + \frac{3}{5^2.7^2} + \frac{4}{7^2.9^2} + ...... + \frac{15}{29^2.31^2}$$ is

from given series we can say that,

$$\ T_n=\frac{n\ }{\left(2n-1\right)^2\left(2n+1\right)^2}=\frac{1\ }{8}\left(\ \frac{\ 1}{\left(2n-1\right)^2}-\frac{1\ }{\left(2n+1\right)^2}\right),\ where\ n=1,2,3,...,15.$$

So,$$S_n=$$

$$\ T_1+T_2+....+T_{15}$$

$$=\frac{1\ }{8}\left(\ \frac{\ 1}{1^2}-\frac{1\ }{3^2}\right)+\frac{1\ }{8}\left(\frac{1\ }{3^2}-\frac{1\ }{5^2}\right)+.....+\frac{1\ }{8}\left(\frac{1\ }{29^2}-\frac{1\ }{31^2}\right).$$

$$=\frac{1\ }{8}\left(1-\frac{1\ }{31^2}\right).$$

$$=\frac{1\ }{8}\left(\ \frac{\ 961-1}{961}\right).$$

$$=\frac{1\ }{8}\left(\ \frac{\ 960}{961}\right).$$

$$=\frac{\ 120}{961}.$$

B is correct choice.