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Ta có :
\(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{2015.2016}\)
\(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{2015}-\frac{1}{2016}\)
\(A=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{2015}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2016}\right)\)
\(A=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{2015}+\frac{1}{2016}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2016}\right)\)
\(A=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{2015}+\frac{1}{2016}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1008}\right)\)
\(\Rightarrow A=\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}\)
Vậy \(B-A=\left(\frac{1}{1008}+\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}\right)-\left(\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}\right)\)
\(\Rightarrow B-A=\frac{1}{1008}\)
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=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{100.101}+\frac{1}{101.102}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{101}-\frac{1}{102}\)
\(A=1-\frac{1}{102}\)
\(A=\frac{101}{102}\)
\(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{2005.2006}\)
\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2005}-\frac{1}{2006}\)
\(=\left(1+\frac{1}{3}+...+\frac{1}{2005}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2006}\right)\)
\(=\left(1+\frac{1}{2}+...+\frac{1}{2006}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2006}\right)\)
\(=\left(1+\frac{1}{2}+...+\frac{1}{2006}\right)-\left(1+\frac{1}{2}+...+\frac{1}{1003}\right)\)
\(=\frac{1}{1004}+\frac{1}{1005}+...+\frac{1}{2006}\)(1)
\(B=\frac{1}{1004.2006}+\frac{1}{1005.2005}+....+\frac{1}{2006.1004}\)
\(\Rightarrow\frac{1}{1004}+\frac{1}{2006}+\frac{1}{1005}+\frac{1}{2005}+...+\frac{1}{2006}+\frac{1}{1004}=2\left(\frac{1}{1004}+\frac{1}{1005}+...+\frac{1}{2006}\right)\)
\(=\frac{\frac{1}{1004}+\frac{1}{1005}+...+\frac{1}{2006}}{1505}\)(2)
Thế (1) và (2) vào ta có:
\(\frac{A}{B}=\frac{\frac{1}{1004}+\frac{1}{1005}+...+\frac{1}{2006}}{\frac{\frac{1}{1004}+\frac{1}{1005}+...+\frac{1}{2006}}{1505}}\)
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=1−12+13−14+...+12005−12006=(1+12+...+12006)−(1+12+..+11003)=11004+11005+...+12006A=1−12+13−14+...+12005−12006=(1+12+...+12006)−(1+12+..+11003)=11004+11005+...+12006
Lại có 13010.B=11004+12006+11005+12005+...+11004=11505(11004+11005+...+12006)13010.B=11004+12006+11005+12005+...+11004=11505(11004+11005+...+12006)
Suy ra A/B = 1505
Tham khảo nha
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{2005.2006}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{2005}-\frac{1}{2006}\)
\(=1-\frac{1}{2006}\)
\(=\frac{2005}{2006}\)
A = \(\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2005}-\frac{1}{2006}\)
=\(\left(\frac{1}{1}+\frac{1}{3}+...+\frac{1}{2005}\right)\)\(-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2006}\right)\)
= \(\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2005}+\frac{1}{2006}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2006}\right)\)
= \(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2005}+\frac{1}{2006}\)\(-\frac{1}{1}-\frac{1}{2}-...-\frac{1}{1003}\)
= \(\frac{1}{1004}+\frac{1}{1005}+...+\frac{1}{2005}+\frac{1}{2006}\)
(=) B - A = \(\frac{1}{1008}+\frac{1}{1009}+...+\frac{1}{2015}+\frac{1}{2016}\)- \(\frac{1}{1004}-\frac{1}{1005}-...-\frac{1}{2005}-\frac{1}{2006}\)
= \(\frac{1}{2007}+\frac{1}{2008}+...+\frac{1}{2016}-\) \(\frac{1}{1004}-\frac{1}{1005}-\frac{1}{1006}-\frac{1}{1007}\)