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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 : \(\frac{9}{10!}+\frac{10}{11!}+\frac{11}{12!}+...+\frac{99}{100!}\)
\(=\frac{10-1}{10!}+\frac{11-1}{11!}+\frac{12-1}{12!}+...+\frac{100-1}{100!}\)
\(=\frac{1}{9!}-\frac{1}{10!}+\frac{1}{10!}-\frac{1}{11!}+\frac{1}{11!}-\frac{1}{12!}+...+\frac{1}{99!}-\frac{1}{100!}\)
\(=\frac{1}{9!}-\frac{1}{100!}< \frac{1}{9!}\left(đpcm\right)\)
\(A=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\)
\(\frac{1}{1^2}=1\)
Ta có :
\(\frac{1}{2^2}< \frac{1}{1\cdot2};\frac{1}{3^2}< \frac{1}{2\cdot3};\frac{1}{4^2}< \frac{1}{3\cdot4}\)
\(1+\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{49\cdot50}\)
\(=1+\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)
\(=1+1-\frac{1}{50}\)
\(=2-\frac{1}{50}\)
\(\Rightarrow A< 2-\frac{1}{50}< 2\left(dpcm\right)\)
Đặt \(B=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{49\cdot50}\)
\(\frac{1}{1^2}< \frac{1}{1\cdot2}\)
\(\frac{1}{2^2}< \frac{1}{2\cdot3}\)
...
\(\frac{1}{50^2}< \frac{1}{49\cdot50}\)
\(B=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)
\(B=1-\frac{1}{50}< 2\)
\(\Rightarrow A< B< 2\)(đpcm)
\(\frac{1}{2^2}< \frac{1}{1}-\frac{1}{2}\)
\(\frac{1}{3^2}< \frac{1}{2}-\frac{1}{3}\)
\(.......\)
\(\frac{1}{50^2}< \frac{1}{49}-\frac{1}{50}\)
\(\Rightarrow A< \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)
\(\Rightarrow A< \frac{1}{1}-\frac{1}{50}=\frac{49}{50}\)
Mà \(\frac{49}{50}< 2\)
\(\Rightarrow A< 2\)
CHO \(A=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{50^2}.\)CHỨNG MINH A<2
\(\frac{1}{2^2}< \frac{1}{1.2}\)
...................\(\frac{1}{50^2}< \frac{1}{49.50}\)
\(\Rightarrow A< \frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{49.50}\)
\(\Rightarrow A< 1-\frac{1}{50}< \frac{49}{50}< 1< 2\)
1/2^2<1/1*2;1/3^2<1/2*3;1/4^2<1/3*4;1/50^2<1/49*50
ta có:
=> 1/1^2+1/2*3+1/3*4+...+1/49*50
<=> 1/1-1/2+1/2-1/3+1/3-1/4+...+1/49-1/50
<=> 1-1/50 < 2
=> A < 2
\(A=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\)
\(A< 1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...\frac{1}{49.50}\)
\(A< 1+\frac{49}{50}\)
\(A< 1\frac{49}{50}\)
Mà \(\frac{49}{50}< 2\)nên 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
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