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A = \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)
A < \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
A < \(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
A < 1 - \(\frac{1.}{100}\)
A < \(\frac{99}{100}< \frac{199}{100}\)
=> A < \(\frac{199}{100}\)
b,
S = \(\frac{3}{2^2}.\frac{8}{3^2}.\frac{15}{4^2}...\frac{99}{10^2}\)
S = \(\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}...\frac{9.11}{10.10}\)
S = \(\frac{1.3.2.4.3.5.4.6.5.7...9.11}{2.2.3.3.4.4...10.10}\)
S = \(\frac{1.2.3^2.4^2.5^2...9^2.10.11}{2^2.3^3.4^2...10^2}\)
S = \(\frac{1.11}{2.10}\)
S = \(\frac{11}{20}\)
A=(\(\frac{1}{2^2}\) -1).( \(\frac{1}{3^2}\)-1)............(\(\frac{1}{100^2}\) -1)=\(-\frac{\left(1.2.3.4....99\right)\left(1.2.3.4....101\right)}{\left(1.2.3.4....100\right)\left(1.2.3.4....100\right)}\)=\(\frac{-101}{100}\)
Ta có :
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\)
\(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\)
\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\right)\)
\(A=1-\frac{1}{2^{100}}< 1\) ( đpcm )
Vậy \(A< 1\)
Chúc bạn học tốt ~
\(\Rightarrow2A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}\)
\(A=1-\frac{1}{2^{99}}< 1\\ \)
Ta có :
\(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)...\left(\frac{1}{100^2}-1\right)\)
\(A=\frac{1-2^2}{2^2}.\frac{1-3^2}{3^2}...\frac{1-100^2}{100^2}\)
\(A=\frac{\left(1-2\right)\left(1+2\right)\left(1-3\right)\left(1+3\right)...\left(1-100\right)\left(1+100\right)}{2^2.3^2...100^2}\)
\(A=\frac{\left(-1\right)\left(-2\right)...\left(-99\right)}{2.3...100}.\frac{3.4...101}{2.3...100}=\frac{-1}{100}.\frac{101}{2}=\frac{-101}{200}< \frac{-100}{200}=\frac{-1}{2}\)
\(\frac{3}{2^2}.\frac{8}{3^2}.\frac{15}{4^2}.....\frac{899}{30^2}\)
\(=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}.....\frac{29.31}{30.30}=\frac{1.2.3.....29}{2.3.4.....30}.\frac{3.4.5.....31}{2.3.4.....30}\)
\(=\frac{1}{2}.\frac{31}{30}=\frac{31}{60}\)
\(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)...\left(\frac{1}{100^2}-1\right)\)
\(\Rightarrow A=\left(\frac{1}{2^2}-\frac{4}{2^2}\right)\left(\frac{1}{3^2}-\frac{9}{3^2}\right)...\left(\frac{1}{100^2}-\frac{10000}{100^2}\right)\)
\(\Rightarrow A=\frac{-3}{2^2}.\frac{-8}{3^2}...\frac{-9999}{100^2}\)
\(\Rightarrow A=-\frac{3}{2^2}.\frac{8}{3^2}...\frac{9999}{100^2}\)
\(\Rightarrow A=-\frac{1.3}{2.2}.\frac{2.4}{3.3}...\frac{99.101}{100.100}\)
\(\Rightarrow A=-\frac{\left(1.2...99\right)\left(3.4...101\right)}{\left(2.3...100\right)\left(2.3...100\right)}\)
\(\Rightarrow A=-\frac{101}{100.2}=\frac{-101}{200}< \frac{-100}{200}=\frac{-1}{2}\)
Vậy \(A< \frac{-1}{2}\)
Ta có :
\(A=\left(\frac{1}{2^2}-1\right).\left(\frac{1}{3^2}-1\right)....\left(\frac{1}{100^2}-1\right)\)
\(A< \left(\frac{1}{2}-1\right).\left(\frac{1}{3}-1\right)...\left(\frac{1}{100}-1\right)\)
\(\Rightarrow A< \left(\frac{-1}{2}\right).\left(\frac{-2}{3}\right)....\left(\frac{-99}{100}\right)\)
\(\Rightarrow A< -\left(\frac{1}{2}.\frac{2}{3}...\frac{99}{100}\right)\)
\(A< -\left(\frac{1.2....99}{2.3...100}\right)=\frac{-1}{100}\)
\(\)Mà \(\frac{-1}{100}>\frac{-1}{2}\)
\(\Rightarrow A>\frac{-1}{2}\)
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