Question

14．计算：$\frac{1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+…+\frac{1}{99}-\frac{1}{100}}{\frac{1}{100×102}+\frac{1}{100×104}+…+\frac{1}{100×200}}$．

Answer 1

分析 原式分子分母变形后，约分即可得到结果．

解答 解：原式=$\frac{1+\frac{1}{3}+\frac{1}{5}+…+\frac{1}{99}-\frac{1}{2}-\frac{1}{4}-…-\frac{1}{100}}{\frac{1}{100}（\frac{1}{102}+\frac{1}{104}+…+\frac{1}{200}）}$
=$\frac{1+\frac{1}{2}+…+\frac{1}{100}-2（\frac{1}{2}+\frac{1}{4}+…+\frac{1}{100}）}{\frac{1}{200}（\frac{1}{51}+\frac{1}{52}+…+\frac{1}{100}）}$
=$\frac{\frac{1}{51}+\frac{1}{52}+…+\frac{1}{100}}{\frac{1}{200}（\frac{1}{51}+\frac{1}{52}+…+\frac{1}{100}）}$
=200．

点评 此题考查了有理数的混合运算，熟练掌握运算法则是解本题的关键．