Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+..............+\frac{1}{99^2}\)
\(A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+................+\frac{1}{98.99}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+............+\frac{1}{98}-\frac{1}{99}\)
\(=1-\frac{1}{99}=\frac{98}{99}< 1\)
\(A>\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+.............+\frac{1}{99.100}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...............+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{2}-\frac{1}{100}=\frac{49}{100}\)
Vậy \(\frac{49}{100}< A< 1\)
\(\frac{3}{2^2}\cdot\frac{8}{3^2}\cdot\frac{15}{4^2}\cdot.....\cdot\frac{899}{30^2}\)
\(=\frac{1\cdot3}{2\cdot2}\cdot\frac{2\cdot4}{3\cdot3}\cdot\frac{3\cdot5}{4\cdot4}\cdot.....\cdot\frac{29\cdot31}{30\cdot30}\)
\(=\frac{1}{2}\cdot\frac{3}{2}\cdot\frac{2}{3}\cdot\frac{4}{3}\cdot\frac{3}{4}\cdot\frac{5}{4}\cdot....\cdot\frac{29}{30}\cdot\frac{31}{30}\)
\(=\left(\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot....\cdot\frac{29}{30}\right)\cdot\left(\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}\cdot....\cdot\frac{31}{30}\right)\)
\(=\frac{1}{30}\cdot\frac{31}{2}\)
\(=\frac{31}{60}\)
b, \(A=\frac{3}{10}+\frac{3}{11}+\frac{3}{12}+\frac{3}{13}+\frac{3}{14}\)
Ta có:
\(\frac{3}{15}< \frac{3}{10}=\frac{3}{10}\)
\(\frac{3}{15}< \frac{3}{11}< \frac{3}{10}\)
\(\frac{3}{15}< \frac{3}{12}< \frac{3}{10}\)
\(\frac{3}{15}< \frac{3}{13}< \frac{3}{10}\)
\(\frac{3}{15}< \frac{3}{14}< \frac{3}{10}\)
\(\Rightarrow\frac{3}{15}+\frac{3}{15}+\frac{3}{15}+\frac{3}{15}+\frac{3}{15}< \frac{3}{10}+\frac{3}{11}+\frac{3}{12}+\frac{3}{13}+\frac{3}{14}< \frac{3}{10}+\frac{3}{10}+\frac{3}{10}+\frac{3}{10}+\frac{3}{10}\)
\(\Rightarrow\frac{3\cdot5}{15}< A< \frac{3\cdot5}{10}\)
\(\Rightarrow1< A< \frac{15}{10}=\frac{3}{2}\)
Mà \(\frac{3}{2}< 2\)
\(\Rightarrow1< A< 2\)
c ,Ta có
\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}-2\cdot\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{25}\right)+\left(\frac{1}{26}+\frac{1}{27}+...+\frac{1}{50}\right)-\left(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{25}\right)\)
\(=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{49}+\frac{1}{50}\)
ta có :
\(\frac{1}{2.3}>\frac{1}{3^2}>\frac{1}{4.3};\frac{1}{3.4}>\frac{1}{4^2}>\frac{1}{4.5}....\)
Tương tự ta sẽ có :
\(\frac{1}{2^2}+\frac{1}{2.3}+.+\frac{1}{99.100}>A>\frac{1}{2^2}+\frac{1}{3.4}+..+\frac{1}{100.101}\)
hay ta có :
\(\frac{1}{4}+\frac{1}{2}-\frac{1}{3}+..+\frac{1}{99}-\frac{1}{100}>A>\frac{1}{2^2}+\frac{1}{3}-\frac{1}{4}+..+\frac{1}{100}-\frac{1}{101}\)
hay \(\frac{1}{4}+\frac{1}{2}-\frac{1}{100}>A>\frac{1}{4}+\frac{1}{3}-\frac{1}{101}\)
hay ta có : \(\frac{1}{4}+\frac{1}{2}>A>\frac{1}{4}+\frac{1}{3}-\frac{31}{300}\Leftrightarrow\frac{3}{4}>A>\frac{12}{25}\)
vậy ta có điều phải chứng minh
B = \(\frac{1}{2.2}+\frac{1}{3.3}+....+\frac{1}{9.9}\)
ta có B > \(\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{9.10}\) ( tự giải thích )
=> B > \(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{9}-\frac{1}{10}=\frac{1}{2}-\frac{1}{10}=\frac{2}{5}\)
=> B > \(\frac{2}{5}\) (1)
Ta có B < \(\frac{1}{1.2}+\frac{1}{2.3}+.....+\frac{1}{8.9}\)
=> B < \(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{8}-\frac{1}{9}=1-\frac{1}{9}=\frac{8}{9}\)
=> B < \(\frac{8}{9}\) (2)
Từ (1) và (2) => \(\frac{2}{5}< B< \frac{8}{9}\)