Lê Quốc Trần Anh Coordinator
02/08/2017 at 13:20-
Dao Trong Luan 02/08/2017 at 13:38\(\left(1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{99}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)\)
\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+...+\dfrac{1}{100}\right)\)
\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{50}\right)\)
\(=\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}\)
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