B=\(\frac{12}{2^{2} . 4^{2}} + \frac{20}{4^{2} . 6^{2}} + . . . . . . + \frac{388}{9 6^{2} . 9 8^{2}} + \frac{396}{9 8^{2} . 10 0^{2}}\)
=\(\frac{1}{2^{2}} - \frac{1}{4^{2}} + \frac{1}{4^{2}} - \frac{1}{6^{2}} + . . . + \frac{1}{9 6^{2}} - \frac{1}{9 8^{2}} + \frac{1}{9 8^{2}} - \frac{1}{10 0^{2}}\)
=\(\frac{1}{2^{2}} - \frac{1}{10 0^{2}}\)
=\(\frac{2599}{10000} < \frac{2500}{10000} = \frac{1}{4}\)
=> B<\(\frac{1}{4}\)
