Lời giải:
$S=3(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{41.42})$
$=3(\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+....+\frac{42-41}{41.42})$
$=3(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{41}-\frac{1}{42})$
$=3(\frac{1}{2}-\frac{1}{42})=\frac{10}{7}$