Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Đề bài : Cho \(A=\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{10^2}\).
Chứng minh : \(\frac{8}{33}< A< \frac{2}{5}\).
Giải : Ta có : \(\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+...+\frac{1}{10\cdot11}< A< \frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{9\cdot10}\)
\(\frac{1}{3}-\frac{1}{11}< A< \frac{1}{2}-\frac{1}{10}\)
\(\frac{1}{11}-\frac{3}{33}=\frac{8}{22}< A< \frac{5}{10}-\frac{1}{10}=\frac{2}{5}\)
\(\frac{8}{33}< A< \frac{2}{5}\)
Ta có: \(A=\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+....+\frac{1}{10^2}< \frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{9\cdot10}\)
\(\Rightarrow A< \frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\)
\(\Rightarrow A< \frac{2}{5}\)(1)
Lại có: \(A=\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{10^2}>\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+...+\frac{1}{10\cdot11}\)
\(\Rightarrow A>\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{10}-\frac{1}{11}\)
\(\Rightarrow A>\frac{8}{33}\)(2)
Từ (1)(2) suy ra \(\frac{8}{33}< A< \frac{2}{5}\)
Vậy...
:
sửa đề : \(\frac{9}{10!}+\frac{10}{11!}+\frac{11}{12!}+...+\frac{99}{100!}\)
\(=\frac{10-1}{10!}+\frac{11-1}{11!}+\frac{12-1}{12!}+...+\frac{100-1}{100!}\)
\(=\frac{1}{9!}-\frac{1}{10!}+\frac{1}{10!}-\frac{1}{11!}+\frac{1}{11!}-\frac{1}{12!}+...+\frac{1}{99!}-\frac{1}{100!}\)
\(=\frac{1}{9!}-\frac{1}{100!}< \frac{1}{9!}\left(đpcm\right)\)
\(A=1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}< 1+\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{49\cdot50}\)
\(=1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}=2-\frac{1}{50}< 2\)
\(\Rightarrow A< 2\)
\(A=1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}.\)
\(A=1+\frac{1}{2\cdot2}+\frac{1}{3\cdot3}+\frac{1}{4\cdot4}+.......+\frac{1}{50\cdot50}\)
\(< 1+\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+....+\frac{1}{49\cdot50}.\)
\(\Rightarrow1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(1+1-\frac{1}{50}< 2\)
=>A<2
ok xong
A = \(\frac{1}{1^2}\) + \(\frac{1}{2^2}\) + \(\frac{1}{3^2}\)+ \(\frac{1}{4^2}\) + .... + \(\frac{1}{50^2}\)
A = 1 + \(\frac{1}{2.2}\)+ \(\frac{1}{3.3}\)+ \(\frac{1}{4.4}\)+ ...... + \(\frac{1}{50.50}\)< 1 + \(\frac{1}{1.2}\)+ \(\frac{1}{2.3}\)+ \(\frac{1}{3.4}\)+ ...... + \(\frac{1}{49.50}\)
A < 1 + ( 1 - \(\frac{1}{2}\)+ \(\frac{1}{2}\)- \(\frac{1}{3}\)+ \(\frac{1}{3}\)- \(\frac{1}{4}\)+ ...... + \(\frac{1}{49}\)- \(\frac{1}{50}\))
A < 1 + ( 1 - \(\frac{1}{50}\))
A < 1 + 1 - \(\frac{1}{50}\)
A < 2 - \(\frac{1}{50}\)
=> A < 2
Bạn xem lời giải ở đường link sau nhé:
Câu hỏi của nguyenducminh - Toán lớp 6 - Học toán với OnlineMath
A=\(\frac{1}{1^2}\)\(+\frac{1}{2^2}\)\(+\frac{1}{3^2}\)\(+...+\frac{1}{50^2}\)
A<1\(+\frac{1}{1.2}\)\(+\frac{1}{2.3}\)\(+...\frac{1}{49.50}\)
=1+1-\(-\frac{1}{2}\)\(+\frac{1}{2}\)\(-\frac{1}{3}\)\(+...+\frac{1}{49}\)\(-\frac{1}{50}\)
=\(1+1-\frac{1}{50}\)
=\(2-\frac{1}{50}\)\(< 2\)
\(\Rightarrow A< 2\)
Ta có : \(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\)\(=1+\frac{1}{2.2}+\frac{1}{3.3}+...+\frac{1}{50.50}\)
Vì \(\frac{1}{2.2}< \frac{1}{1.2};\frac{1}{3.3}< \frac{1}{2.3};..;\frac{1}{50.50}< \frac{1}{49.50}\)nên :
\(\Rightarrow\) \(1+\frac{1}{2.2}+\frac{1}{3.3}+...+\frac{1}{50.50}\)\(< 1+\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{49.50}\)
Ta có : \(1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\)
\(=1+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\right)\)
\(=1+\left(1-\frac{1}{50}\right)\)\(=1+\frac{49}{50}\)
Vì \(\frac{49}{50}< 1\)nên \(1+\frac{49}{50}< 2\)\(\Rightarrow\)\(1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}< 2\)
\(\Rightarrow\)\(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\)\(< 1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}< 2\)
Ta có :
\(\frac{50}{111}>\frac{50}{200}\)
\(\frac{50}{112}>\frac{50}{200}\)
\(\frac{50}{113}>\frac{50}{200}\)
\(\frac{50}{114}>\frac{50}{200}\)
\(\Rightarrow A>\frac{50}{200}+\frac{50}{200}+\frac{50}{200}+\frac{50}{200}\)hay \(A>\frac{50}{200}.4\left(1\right)\)
Mặt khác :
\(\frac{50}{111}< \frac{50}{100}\)
\(\frac{50}{112}< \frac{50}{100}\)
\(\frac{50}{113}< \frac{50}{100}\)
\(\frac{50}{114}< \frac{50}{100}\)
\(\Rightarrow A< \frac{50}{100}+\frac{50}{100}+\frac{50}{100}+\frac{50}{100}\)hay \(A< \frac{50}{100}.4\left(2\right)\)
Từ \(\left(1\right)\)và \(\left(2\right)\Rightarrow1< A< 2\left(đpcm\right)\)
A = 1/12 + 1/22 + 1/32 + ... + 1/502
A = 1/1.1 + 1/2.2 + 1/3.3 + ... + 1/50.50
A < 1/1 + 1/1.2 + 1/2.3 + ... + 1/49.50
A < 1 + 1 - 1/2 + 1/2 - 1/3 + ... + 1/49 - 1/50
A < 2 - 1/50 < 2
Chứng tỏ A < 2
Đặt B=1/1+1/1.2+...+1/49.50
Ta có:
A=1/1^2+1/2^2+...+1/50^2<B=1/1+1/1.2+...+1/49.50 (1)
Mà B=1/1+1/1.2+...+1/49.50
=1+1-1/2+1/2-1/3+...+1/49-1/50
=2-1/50 <2 (2)
Từ (1) và (2) =>A<B<2
=>A<2