\(M=\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{2003.2005}\)
\(=\dfrac{1}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{2003.2005}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2003}-\dfrac{1}{2005}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{2005}\right)\)
\(=\dfrac{1}{2}.\dfrac{2004}{2005}=\dfrac{1002}{2005}\)
2M= 1/1.3+1/3.5+1/5.7+...+1/2003.2005
2M= 1/1-1/3+1/3-1/5+...+1/2003-1/2005
2M= 1/1-1/2005
2M= 2004/2005
M= 2004/2005:2
M=1002/2005
\(M=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{2003.2005}\)
= \(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2003}-\dfrac{1}{2005}\)
= \(1-\dfrac{1}{2005}\)
= \(\dfrac{2004}{2005}\)
\(M=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{2003.2005}\)
Ta thấy :\(\dfrac{1}{1.3}=1-\dfrac{1}{3};\dfrac{1}{3.5}=\dfrac{1}{3}-\dfrac{1}{5};\dfrac{1}{5.7}=\dfrac{1}{5}-\dfrac{1}{7};\)
...;\(\dfrac{1}{2003.2005}=\dfrac{1}{2003}-\dfrac{1}{2005}\)
\(\Rightarrow M=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\)
\(\dfrac{1}{2003}-\dfrac{1}{2005}\)
\(M=1-\dfrac{1}{2005}=\dfrac{2005}{2005}-\dfrac{1}{2005}=\dfrac{2004}{2005}\)
Vậy \(M=\dfrac{2004}{2005}\)
