Question

19．當(dāng)n是正整數(shù)時(shí)，因?yàn)椋╪+1）²＞n（n+2）所以有$\frac{n}{n+1}$＜$\frac{n+1}{n+2}$，參考這個(gè)結(jié)論，比較A=$\frac{1}{2}$×$\frac{3}{4}$×$\frac{5}{6}$×…×$\frac{99}{100}$與B=$\frac{1}{10}$的大小A＜B．

Answer 1

分析 根據(jù)題意可得，$\frac{1}{2}＜\frac{2}{3}$，$\frac{5}{6}＜\frac{6}{7}$，…，將式子兩個(gè)一組進(jìn)行比較，依此繼續(xù)下去，就可以得到$\frac{1}{2}$×$\frac{3}{4}$×$\frac{5}{6}$×…×$\frac{99}{100}$＜$\frac{2}{3}×\frac{4}{5}×\frac{6}{7}×$…×$\frac{98}{99}$，最后兩邊同時(shí)乘以$\frac{1}{2}$×$\frac{3}{4}$×$\frac{5}{6}$×…×$\frac{99}{100}$得到問題的答案．

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．

點(diǎn)評 本題考查對分式乘法的掌握，關(guān)鍵是找出規(guī)律，進(jìn)行化簡．