解：（1）\(\because \)\(f(1) = 1 - \frac{1}{2}\)，\(f(2) = 1 - \frac{1}{3}\)，\(f(3) = 1 - \frac{1}{4}\)，\(f(4) = 1 - \frac{1}{5}\)，\( \ldots \)

\(\therefore f(n) = 1 - \frac{1}{{n + 1}}\)．

\(f\)（1）\( \cdot f\)（2）\( \cdot f\)（3）\( \cdot \ldots \cdot f(100)\)

\( = (1 - \frac{1}{2})(1 - \frac{1}{3})(1 - \frac{1}{4}) \cdot \cdot \cdot (1 - \frac{1}{{101}})\)

\( = \frac{1}{2} \times \frac{2}{3} \times \frac{3}{4} \times \cdot \cdot \cdot \times \frac{{100}}{{101}}\)

\( = \frac{1}{{101}}\)．

故答案为：\(1 - \frac{1}{{n + 1}}\)；\(\frac{1}{{101}}\)．

进入考试题库