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\(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{2013.2014}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{2013}-\frac{1}{2014}\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{2013}+\frac{1}{2014}-2.\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2014}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{2013}+\frac{1}{2014}-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1007}\right)\)
\(=\frac{1}{1008}+\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2013}+\frac{1}{2014}\)
Lại có B = \(\frac{1}{1008.2014}+\frac{1}{1009.2013}+\frac{1}{1010.2012}+...+\frac{1}{2014.1008}\)
=> 3022B = \(\frac{3022}{1008.2014}+\frac{3022}{1009.2013}+\frac{3022}{1010.2012}+...+\frac{3022}{2014.1008}\)
\(=\frac{1}{1008}+\frac{1}{2014}+\frac{1}{1009}+\frac{1}{2013}+\frac{1}{1010}+\frac{1}{2012}+...+\frac{1}{2014}+\frac{1}{1008}\)
\(=2.\left(\frac{1}{1008}+\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2013}+\frac{1}{2014}\right)\)
=> \(B=\frac{1}{1511}.\left(\frac{1}{1008}+\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2013}+\frac{1}{2014}\right)\)
Khi đó \(\frac{A}{B}=\frac{\left(\frac{1}{1008}+\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2013}+\frac{1}{2014}\right)}{\frac{1}{1511}.\left(\frac{1}{1008}+\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2013}+\frac{1}{2014}\right)}=\frac{1}{\frac{1}{1511}}=1511\)
=> \(\frac{A}{B}=1511\)
=> A/B là 1 số nguyên (đpcm)
\(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{2017}-\frac{1}{2018}\)
\(A=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{2017}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2018}\right)\)
\(A=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}\right)-2\cdot\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2018}\right)\)\(A=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}-1-\frac{1}{2}-...-\frac{1}{1009}\)
\(A=\frac{1}{1010}+\frac{1}{2000}+...+\frac{1}{2018}\)
\(B=3028.\left(\frac{1}{1010.2018}+...+\frac{1}{2018.1010}\right)\)
\(B=\frac{3028}{1010.2018}+...+\frac{3028}{2018.1010}\)
\(B=\frac{1}{1010}+\frac{1}{2018}+...+\frac{1}{2018}+\frac{1}{1010}\)
\(B=2.\left(\frac{1}{1010}+...+\frac{1}{2018}\right)\)
\(=>\frac{A}{B}=\frac{1}{2}\)
Linh Phương Ngô chứng minh a/b là số nguyên cơ mà