Câu hỏi của ๖ۣۜBá ๖ۣۜVươηɠ

Question

Chứng minh rằng:$\frac{7}{12}< \frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\frac{1}{5\cdot6}+...+\frac{1}{99\cdot100}< \frac{5}{6}$

kudo shinichi · Accepted Answer

$\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{99.100}$$=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}$$=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)$$=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{99}+\frac{1}{100}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)$$=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{99}+\frac{1}{100}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)$$=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}$Ta có:$\frac{1}{51}>\frac{1}{75}$$\frac{1}{52}>\frac{1}{75}$......................$\frac{1}{75}=\frac{1}{75}$$\Rightarrow\frac{1}{51}+\frac{1}{52}+...+\frac{1}{75}>\frac{1}{75}+\frac{1}{75}+...+\frac{1}{75}=25.\frac{1}{75}=\frac{1}{3}$(1)Ta có:$\frac{1}{76}>\frac{1}{100}$$\frac{1}{77}>\frac{1}{100}$........................$\frac{1}{100}=\frac{1}{100}$$\Rightarrow\frac{1}{76}+\frac{1}{77}+...+\frac{1}{100}>\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}=25.\frac{1}{100}=\frac{1}{4}$(2)Từ (1) và (2) ta có:$\frac{1}{51}+\frac{1}{52}+...+\frac{1}{75}+\frac{1}{76}+...+\frac{1}{100}>\frac{1}{3}+\frac{1}{4}=\frac{7}{12}$$\Rightarrow\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}>\frac{7}{12}$(5)Ta có:$\frac{1}{51}=\frac{1}{51}$$\frac{1}{52}< \frac{1}{51}$...................$\frac{1}{75}< \frac{1}{51}$$\Rightarrow\frac{1}{51}+\frac{1}{52}+...+\frac{1}{75}< \frac{1}{51}+\frac{1}{51}+...+\frac{1}{51}=25.\frac{1}{51}< 25.\frac{1}{50}=\frac{1}{2}$(3)Ta có:$\frac{1}{76}=\frac{1}{76}$$\frac{1}{77}< \frac{1}{76}$...................$\frac{1}{100}< \frac{1}{76}$$\Rightarrow\frac{1}{76}+\frac{1}{77}+...+\frac{1}{100}< \frac{1}{76}+\frac{1}{76}+...+\frac{1}{76}=25.\frac{1}{76}< 25.\frac{1}{75}=\frac{1}{3}$(4)Từ (3) và (4) ta có:$\frac{1}{51}+\frac{1}{52}+...+\frac{1}{75}+\frac{1}{76}+...+\frac{1}{100}>\frac{1}{2}+\frac{1}{3}=\frac{5}{6}$$\Rightarrow\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}< \frac{5}{6}$(6)Từ (5) và (6) $\Rightarrow\frac{7}{12}< \frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{99.100}< \frac{5}{6}$ đpcmTham khảo nhé~Câu trả lời: $\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{99.100}$$=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}$$=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)$$=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{99}+\frac{1}{100}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)$$=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{99}+\frac{1}{100}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)$$=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}$Ta có:$\frac{1}{51}>\frac{1}{75}$$\frac{1}{52}>\frac{1}{75}$......................$\frac{1}{75}=\frac{1}{75}$$\Rightarrow\frac{1}{51}+\frac{1}{52}+...+\frac{1}{75}>\frac{1}{75}+\frac{1}{75}+...+\frac{1}{75}=25.\frac{1}{75}=\frac{1}{3}$(1)Ta có:$\frac{1}{76}>\frac{1}{100}$$\frac{1}{77}>\frac{1}{100}$........................$\frac{1}{100}=\frac{1}{100}$$\Rightarrow\frac{1}{76}+\frac{1}{77}+...+\frac{1}{100}>\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}=25.\frac{1}{100}=\frac{1}{4}$(2)Từ (1) và (2) ta có:$\frac{1}{51}+\frac{1}{52}+...+\frac{1}{75}+\frac{1}{76}+...+\frac{1}{100}>\frac{1}{3}+\frac{1}{4}=\frac{7}{12}$$\Rightarrow\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}>\frac{7}{12}$(5)Ta có:$\frac{1}{51}=\frac{1}{51}$$\frac{1}{52}< \frac{1}{51}$...................$\frac{1}{75}< \frac{1}{51}$$\Rightarrow\frac{1}{51}+\frac{1}{52}+...+\frac{1}{75}< \frac{1}{51}+\frac{1}{51}+...+\frac{1}{51}=25.\frac{1}{51}< 25.\frac{1}{50}=\frac{1}{2}$(3)Ta có:$\frac{1}{76}=\frac{1}{76}$$\frac{1}{77}< \frac{1}{76}$...................$\frac{1}{100}< \frac{1}{76}$$\Rightarrow\frac{1}{76}+\frac{1}{77}+...+\frac{1}{100}< \frac{1}{76}+\frac{1}{76}+...+\frac{1}{76}=25.\frac{1}{76}< 25.\frac{1}{75}=\frac{1}{3}$(4)Từ (3) và (4) ta có:$\frac{1}{51}+\frac{1}{52}+...+\frac{1}{75}+\frac{1}{76}+...+\frac{1}{100}>\frac{1}{2}+\frac{1}{3}=\frac{5}{6}$$\Rightarrow\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}< \frac{5}{6}$(6)Từ (5) và (6) $\Rightarrow\frac{7}{12}< \frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{99.100}< \frac{5}{6}$ đpcmTham khảo nhé~

Khoá học trên OLM (olm.vn)

Khoá học trên OLM (olm.vn)