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- Hướng dẫn giải

$\begin{array}{l}B=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+\mathrm{...}+\frac{3}{97.100}\\ =1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}\mathrm{...}+\frac{1}{97}-\frac{1}{100}\\ =1-\frac{1}{100}=\frac{99}{100}\end{array}$