Question

19．当n是正整数时，因为（n+1）²＞n（n+2）所以有$\frac{n}{n+1}$＜$\frac{n+1}{n+2}$，参考这个结论，比较A=$\frac{1}{2}$×$\frac{3}{4}$×$\frac{5}{6}$×…×$\frac{99}{100}$与B=$\frac{1}{10}$的大小A＜B．

Answer 1

分析 根据题意可得，$\frac{1}{2}＜\frac{2}{3}$，$\frac{5}{6}＜\frac{6}{7}$，…，将式子两个一组进行比较，依此继续下去，就可以得到$\frac{1}{2}$×$\frac{3}{4}$×$\frac{5}{6}$×…×$\frac{99}{100}$＜$\frac{2}{3}×\frac{4}{5}×\frac{6}{7}×$…×$\frac{98}{99}$，最后两边同时乘以$\frac{1}{2}$×$\frac{3}{4}$×$\frac{5}{6}$×…×$\frac{99}{100}$得到问题的答案．

解答 解：∵A=$\frac{1}{2}$×$\frac{3}{4}$×$\frac{5}{6}$×…×$\frac{99}{100}$
∴A=$\frac{1}{2}$×$\frac{3}{4}$×$\frac{5}{6}$×…×$\frac{99}{100}$＜$\frac{2}{3}×\frac{4}{5}×\frac{6}{7}×$…×$\frac{98}{99}$
∴（$\frac{1}{2}$×$\frac{3}{4}$×$\frac{5}{6}$×…×$\frac{99}{100}$）×（$\frac{1}{2}$×$\frac{3}{4}$×$\frac{5}{6}$×…×$\frac{99}{100}$）＜（$\frac{2}{3}×\frac{3}{4}$）×$（\frac{6}{7}×\frac{7}{8}）$×…×$（\frac{98}{99}×\frac{99}{100}）$=$\frac{1}{100}$，
∴$\frac{1}{2}$×$\frac{3}{4}$×$\frac{5}{6}$×…×$\frac{99}{100}$＜$\frac{1}{10}$，
∴A＜B．

点评 本题考查对分式乘法的掌握，关键是找出规律，进行化简．