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\(M=\frac{1}{2}-\frac{1}{2^4}+\frac{1}{2^7}-\frac{1}{2^{10}}+....+\frac{1}{2^{43}}-\frac{1}{2^{46}}+\frac{1}{2^{49}}-\frac{1}{2^{52}}\)
Nên \(2^3.M=4-\frac{1}{2}+\frac{1}{2^4}-\frac{1}{2^7}+.....+\frac{1}{2^{46}}-\frac{1}{2^{52}}\)
Suy ra \(2^3.M-M=4-\frac{1}{2^{52}}\)hay\(7.M=4-\frac{1}{2^{52}}\).
Khi đó \(M=\frac{4}{7}-\frac{1}{2^{52}.7}< 1\)
Vì \(\frac{9}{4}>1;M< 1\)nên \(\frac{9}{4}>M\)
Vậy \(\frac{9}{4}>M\)
Ta có: \(\frac{3}{1^2.2^2}=\frac{3}{1.4}=1-\frac{1}{4}\); \(\frac{5}{2^2.3^2}=\frac{5}{4.9}=\frac{1}{4}-\frac{1}{9}\); \(\frac{7}{3^2.4^2}=\frac{7}{9.16}=\frac{1}{9}-\frac{1}{16}\); ...; \(\frac{39}{19^2.20^2}=\frac{39}{361.400}=\frac{1}{361}-\frac{1}{400}\)
Gọi tổng đó là A => A=\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+...+\frac{1}{361}-\frac{1}{400}\)
=> \(A=1-\frac{1}{400}=\frac{399}{400}< \frac{400}{400}=1\)
=> A < 1
\(M=\frac{1}{1.2}+\frac{2}{1.2.3}+.....+\frac{9}{1.2.3.....10}\)
\(M=\frac{2-1}{1.2}+\frac{3-1}{1.2.3}+....+\frac{10-1}{1.2......10}\)
\(M=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{6}+....+\frac{10}{1.2.....10}-\frac{1}{1.2.....10}\)
\(M=1-\frac{1}{1.2.3......10}\)
\(M=1-\frac{1}{3628800}\)
Vì \(1=1\)mà \(\frac{1}{3628800}< 1\)nên \(1-\frac{1}{3628800}< 1\)
Vậy \(M< 1\)
Nhận xét: \(\frac{1}{5}< \frac{1}{42};\frac{1}{9}< \frac{1}{42};\frac{1}{10}< \frac{1}{42};\frac{1}{40}< \frac{1}{42}\)
\(\Rightarrow S< \frac{1}{42}+\frac{1}{42}+\frac{1}{42}+\frac{1}{42}+\frac{1}{42}\)
\(\Rightarrow S< \frac{5}{42}< \frac{21}{42}=\frac{1}{2}\)
Vậy S < 1/2
\(=\frac{\frac{6}{5}:\left(\frac{6}{5}-\frac{5}{4}\right)}{\frac{8}{25}+\frac{2}{25}}+\frac{\frac{2027}{25}:\frac{9}{4}}{\left(\frac{38}{7}-\frac{9}{4}\right):\frac{267}{56}}\)
\(=\frac{\frac{6}{5}:\left(\frac{-1}{20}\right)}{\frac{2}{5}}+\frac{\frac{8180}{225}}{\frac{89}{28}:\frac{167}{56}}\)
\(=\frac{-12}{5}:\frac{2}{5}+\frac{8180}{225}:\frac{178}{167}\)
\(=-1+...\)ra số to vcl
Đề sai à ???