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L
18 tháng 1 2020
\(a)A=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2^{100}-1}\)
\(\Rightarrow A=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{2^2}+...+\frac{1}{2^{100}}+\frac{1}{2^{100}-1}\)
\(\Rightarrow A=1+\left(\frac{1}{2}+\frac{1}{3}\right)+\left(\frac{1}{2^2}+...+\frac{1}{7}\right)+\left(\frac{1}{2^3}+...+\frac{1}{15}\right)+...+\left(\frac{1}{2^{99}}+...+\frac{1}{2^{100}-1}\right)\)
\(\Rightarrow A< 1+\frac{1}{2}.2+\frac{1}{4}.4+\frac{1}{8}.8+...+\frac{1}{2^{99}}.2^{99}\)
\(\Rightarrow A< 100\left(đpcm\right)\)
\(b)A=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2^{100}-1}\)
\(\Rightarrow A=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{2^2}+...+\frac{1}{2^{100}}+\frac{1}{2^{100}-1}+\frac{1}{2^{100}}-\frac{1}{2^{100}}\)
\(\Rightarrow A=1+\frac{1}{2}+\left(\frac{1}{3}+\frac{1}{2^2}\right)+\left(\frac{1}{5}+\frac{1}{2^3}\right)+...+\left(\frac{1}{2^{99}+1}+...+\frac{1}{2^{100}}\right)-\frac{1}{2^{100}}\)
\(\Rightarrow A>1+\frac{1}{2}+\frac{1}{2^2}.2+\frac{1}{2^3}.2^2+...+\frac{1}{2^{100}}.2^{99}-\frac{1}{2^{100}}\)
\(\Rightarrow A>1+\frac{1}{2}.100-\frac{1}{2^{100}}\)
\(\Rightarrow A>51-\frac{1}{2^{100}}>51-1\)
\(\Rightarrow A>50\left(đpcm\right)\)
5 tháng 2 2016
Ta có : \(\frac{1}{3^2}<\frac{1}{2.3}=\frac{1}{2}-\frac{1}{3}\)
\(\frac{1}{4^2}<\frac{1}{3.4}=\frac{1}{3}-\frac{1}{4}\)
......
\(\frac{1}{50^2}<\frac{1}{49.50}=\frac{1}{49}-\frac{1}{50}\)
\(\Rightarrow A=\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{50^2}<\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}>\frac{1}{4}\)
\(\Rightarrow A=\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{50^2}<\frac{12}{25}>\frac{1}{4}\)
Vậy \(A>\frac{1}{4}\)
Ý b làm tương tự
28 tháng 4 2018
\(A=1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}< 1+\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{49\cdot50}\)
\(=1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}=2-\frac{1}{50}< 2\)
\(\Rightarrow A< 2\)
NN
15 tháng 3 2017
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 \(1+\frac{49}{50}< 2\)\(\Rightarrow\)\(1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}< 2\)
\(\Rightarrow\)\(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\)\(< 1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}< 2\)