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Câu 1:
B = \(\frac{2999}{1}+\frac{2998}{2}+\frac{2997}{3}+...+\frac{1}{2999}\)
= \(\frac{3000-1}{1}+\frac{3000-2}{2}+\frac{3000-3}{3}+...+\frac{3000-2999}{2999}\)
= \(\left(\frac{3000}{1}+\frac{3000}{2}+\frac{3000}{3}+...+\frac{3000}{2999}\right)-\left(\frac{1}{1}+\frac{2}{2}+\frac{3}{3}+...+\frac{2999}{2999}\right)\)
= \(3000+3000.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2999}\right)-2999\)
= \(3000\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2999}\right)+\frac{3000}{3000}\)
= \(3000\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{3000}\right)\)
\(\Rightarrow\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{3000}}{3000\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{3000}\right)}=\frac{1}{3000}\)
(1+2012/1)(1+2012/2)(1+2012/3).......(1+2012/1000)
(1+1000/1)(1+1000/2)..........(1+1000/2012)
Tính A
\(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{2013.2014}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{2013}-\frac{1}{2014}\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{2013}+\frac{1}{2014}-2.\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2014}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{2013}+\frac{1}{2014}-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1007}\right)\)
\(=\frac{1}{1008}+\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2013}+\frac{1}{2014}\)
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)