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S = \(\frac{1}{20}+\frac{1}{21}...+\frac{1}{199}+\frac{1}{200}\) ( có 181 phân số )
=> S > \(\frac{1}{200}+\frac{1}{200}+...+\frac{1}{200}+\frac{1}{200}\)
=> S > \(\frac{1}{200}.181\)
=> S > \(\frac{181}{200}\)> \(\frac{180}{200}\)= \(\frac{9}{10}\)
Vậy S > 9 / 10
a, Ta có: \(\frac{1}{2^2}>\frac{1}{2.3};\frac{1}{3^2}>\frac{1}{3.4};...;\frac{1}{10^2}>\frac{1}{10.11}\)
\(\Rightarrow S>\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10.11}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}=\frac{1}{2}-\frac{1}{11}=\frac{9}{22}\)
Vậy S > 9/22
b, Ta có: \(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};...;\frac{1}{10^2}< \frac{1}{9.10}\)
\(\Rightarrow S>\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{10.11}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}=1-\frac{1}{10}=\frac{9}{10}\)
Vậy S > 9/10
\(N=\frac{1}{21}+\frac{1}{22}+\frac{1}{23}+\frac{1}{24}+\frac{1}{25}+\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+\frac{1}{29}+\frac{1}{30}\) >\(\frac{1}{30}+\frac{1}{30}+\frac{1}{30}+\frac{1}{30}+\frac{1}{30}+\frac{1}{30}+\frac{1}{30}+\frac{1}{30}+\frac{1}{30}+\frac{1}{30}=\frac{1}{30}.10=\frac{1}{3}\)
=> N > \(\frac{1}{3}\)
\(\frac{1}{4}+\frac{1}{20}+\frac{1}{21}+\frac{1}{22}+\frac{1}{74}+\frac{1}{75}+\frac{1}{76}\)= 0,4330783347
\(\frac{1007}{2013}\)= 0, 5002483855
Vậy :\(\frac{1}{4}+\frac{1}{20}+\frac{1}{21}+\frac{1}{22}+\frac{1}{74}+\frac{1}{75}+\frac{1}{76}\) < \(\frac{1007}{2013}\)
=> C > 1/200 + 1/200 + 1/200 + ...... + 1/200 ( 181 phân số )
=> C > 181/200 > 180/200 = 9/10
<=> C > 9/10