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\(=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)
\(=1-\frac{1}{50}\)\(<1\)
\(\Rightarrow\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}<1\)
Vậy \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}<1\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(=1-\frac{1}{50}\)
\(=\frac{49}{50}\)
\(\Rightarrow\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}<1\)
\(A=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}\\ A< \frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+...+\frac{1}{50\times51}\\ A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{50}-\frac{1}{51}\\ A< 1-\frac{1}{51}=\frac{49}{51}\\ \Rightarrow A< 2\)
A=1 - 1/2 + 1/3 - 1/4 +..+ 1/49 - 1/50
A= 1-( 1/2 + 1/3 ) - ( 1/4 + 1/5 ) -.....-(1/48 + 1/49) - 1/50
A=1 - 5/6 - 9/20 -.....-97/2352 - /150
A= 1 -............cho con lai tu lam nha
\(A=\frac{1}{1^2}+\frac{1}{2^2}+...+\frac{1}{49^2}+\frac{1}{50^2}.\)
\(\Rightarrow A< \frac{1}{1^2}+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{48.49}+\frac{1}{49.50}\)
\(A< 1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...-\frac{1}{49}+\frac{1}{50}\)
\(A< 1+1-\frac{1}{50}=2-\frac{1}{50}< 2\)
\(\Rightarrow A< 2\left(đpcm\right)\)
Ta có :
\(A=\dfrac{50}{111}+\dfrac{50}{112}+\dfrac{50}{113}+\dfrac{50}{114}\)
Ta thấy :
\(\dfrac{50}{111}>\dfrac{50}{200}\)
\(\dfrac{50}{112}>\dfrac{50}{200}\)
\(\dfrac{50}{113}>\dfrac{50}{200}\)
\(\dfrac{50}{114}>\dfrac{50}{200}\)
\(\Rightarrow A>\dfrac{50}{200}+\dfrac{50}{200}+\dfrac{50}{200}+\dfrac{50}{200}\)
\(\Rightarrow A>\dfrac{50}{200}.4=1\) \(\left(1\right)\)
Mặt khác :
\(\dfrac{50}{111}< \dfrac{50}{100}\)
\(\dfrac{50}{112}< \dfrac{50}{100}\)
\(\dfrac{50}{113}< \dfrac{50}{100}\)
\(\dfrac{50}{114}< \dfrac{50}{100}\)
\(\Rightarrow A< \dfrac{50}{100}+\dfrac{50}{100}+\dfrac{50}{100}+\dfrac{50}{100}\)
\(\Rightarrow A< \dfrac{50}{100}.4=2\) \(\left(2\right)\)
Từ \(\left(1\right)+\left(2\right)\Rightarrow1< A< 2\rightarrowđpcm\)