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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)
bài 2:
a)\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{999}-\frac{1}{1000}\)
\(=1-\frac{1}{1000}\)
\(=\frac{999}{1000}\)
mk ko biết bn có sai đề ko nhưng mk chỉ lm theo ý mk hiểu thôi! sai thì thôi nha!
bn làm như vầy nè
a=1/51+1/52+...+1/100
A=1/3.1/7 + 1/2.1/26+....1/2.1/50
A=1/3-1/7+1/2-1/26+...1/2-1/50
A=1/3-1/50
A=47/50
như vầy đó bn tin mik đi
A= 1/1-1/2+1/2-1/3+1/4-1/5+...+1/101-1/102
A=1-1/102=102/102-1/102=101/102
ý b thì chờ mình tí tìm cách lập luận đã nhé
A=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{100.101}+\frac{1}{101.102}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{101}-\frac{1}{102}\)
\(A=1-\frac{1}{102}\)
\(A=\frac{101}{102}\)
\(A=\frac{1}{2.2}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\)
\(A=\frac{1}{4}+\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\right)\)
\(A=\frac{1}{4}+\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\right)\)
( gạch bỏ các phân số giống nhau)
\(A=\frac{1}{4}+\left(\frac{1}{3}-\frac{1}{9}\right)\)
\(A=\frac{1}{4}+\frac{2}{9}\)
\(A=\frac{17}{36}\)
phần b, c bn lm tương tự như phần a nha