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Chứng minh rằng \(\frac{7}{12}<\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{40}<\frac{5}{6}\)
a) \(\frac{1}{2}< \frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}< 1\)
\(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}=\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{60}\right)+\left(\frac{1}{61}+\frac{1}{62}+...+\frac{1}{70}\right)+...+\left(\frac{1}{91}+\frac{1}{92}+...+\frac{1}{100}\right)\)\(\frac{1}{60}\cdot10< \frac{1}{51}+\frac{1}{52}+...+\frac{1}{60}< \frac{1}{50}\cdot10\)
\(\frac{1}{6}< \frac{1}{51}+\frac{1}{52}+...+\frac{1}{60}< \frac{1}{5}\)(1)
\(\frac{1}{70}\cdot10< \frac{1}{61}+\frac{1}{62}+...+\frac{1}{70}< \frac{1}{60}\cdot10\)
\(\frac{1}{7}< \frac{1}{61}+\frac{1}{62}+...+\frac{1}{70}< \frac{1}{6}\)(2)
.... (tương tự )
\(\frac{1}{100}\cdot10< \frac{1}{91}+\frac{1}{92}+...+\frac{1}{100}< \frac{1}{90}\cdot10\)
\(\frac{1}{10}< \frac{1}{91}+...+\frac{1}{100}< \frac{1}{9}\)
Từ (1)(2)(3)(4) và (5)
\(\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}< \frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}< \frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}\)
\(\frac{1}{2}< \frac{1624}{2520}< \frac{1}{51}+...+\frac{1}{100}\)
\(1>\frac{1879}{2520}>\frac{1}{51}+...+\frac{1}{100}\)
Ta có: \(\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)\)
\(=\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}\)(đpcm)
\(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}=\left(1+\frac{1}{2}+...+\frac{1}{100}\right)-\left(1+\frac{1}{2}+...+\frac{1}{50}\right)\)
\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)+\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(\Rightarrowđpcm\)
Ta có :
\(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{99.100}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}+\frac{1}{100}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}+\frac{1}{100}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)\)
\(=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)
ta có:\(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}=\)\(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)\)
=\(\left(1+\frac{1}{3}+...+\frac{1}{99}\right)+\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\) \(-\left(1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{50}\right)\)
=\(\left(1+\frac{1}{3}+...+\frac{1}{99}\right)+\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)-\) \(2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
=\(\left(1+\frac{1}{3}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)