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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(B=\frac{4^2-4^2}{\left(2\cdot4^2\right)}+\frac{6^2-4^2}{4^2\cdot4^2}+.....+\frac{98^2-96^2}{^{ }96^2\cdot98^2}+\frac{ }{ }\)\(\frac{100^2-98^2}{98^2\cdot100^2}\)
\(=\frac{1}{2^2}-\frac{1}{4^2}+\frac{1}{4^2}-\frac{1}{6^2}+.....+\frac{1}{96^2}-\frac{1}{98^2}-\)\(\frac{1}{100^2}\)
\(=\frac{1}{4}-\frac{1}{100^2}< \frac{1}{4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
B=\(\frac{12}{2^2.4^2}+\frac{20}{4^2.6^2}+......+\frac{388}{96^2.98^2}+\frac{396}{98^2.100^2}\)
=\(\frac{1}{2^2}-\frac{1}{4^2}+\frac{1}{4^2}-\frac{1}{6^2}+...+\frac{1}{96^2}-\frac{1}{98^2}+\frac{1}{98^2}-\frac{1}{100^2}\)
=\(\frac{1}{2^2}-\frac{1}{100^2}\)
=\(\frac{2599}{10000}< \frac{2500}{10000}=\frac{1}{4}\)
=> B<\(\frac{1}{4}\)