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![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(1\cdot3\cdot5\cdot9=\frac{1\cdot2\cdot3\cdot4\cdot...\cdot99\cdot100}{2\cdot4\cdot6\cdot...\cdot100}=\frac{1\cdot2\cdot3\cdot4\cdot...\cdot100}{2\cdot1\cdot2\cdot2\cdot...\cdot2\cdot50}=\frac{1\cdot2\cdot3\cdot4\cdot...\cdot100}{1\cdot2\cdot3\cdot...\cdot50\cdot2\cdot2\cdot2\cdot...\cdot2\cdot2}\)
\(=\frac{51\cdot52\cdot...\cdot100}{2\cdot2\cdot2\cdot...\cdot2\cdot2}\)( 50 THỪA SỐ 2 ) \(=\frac{51}{2}\cdot\frac{52}{2}\cdot\frac{53}{2}\cdot...\cdot\frac{100}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\frac{1}{5\cdot6}+...+\frac{1}{99\cdot100}\)
\(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)
\(A=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)
\(A=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}-1-\frac{1}{2}-\frac{1}{3}-...-\frac{1}{50}\)
\(A=\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{1}{2!}+\frac{2!}{4!}+...+\frac{198!}{200!}=\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{199.200}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-...+\frac{1}{199}-\frac{1}{200}=\left(\frac{1}{1}+\frac{1}{2}+...+\frac{1}{200}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{200}\right)\)
\(=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Easy mà =)))
Ta thấy: \(\frac{1}{50}>\frac{1}{100}\); \(\frac{1}{51}>\frac{1}{100}\);....;\(\frac{1}{99}>\frac{1}{100}\)
Mà từ 50 - 99 có 50 số nên ta có 50 phân số 100
Cộng theo từng vế,ta được:
\(S=\frac{1}{50}+\frac{1}{51}+...+\frac{1}{99}>\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}=\frac{50}{100}=\frac{1}{2}^{\left(đpcm\right)}\) (do có 50 phân số 1/100)
nhan hộ mình bn ơi ?