= 1/2 . (1/1 - 1/3 + 1/3 - 1/5 +... + 1/2009 - 1/2011)

= 1/2 . (1/1 - 1/2011)

= 1/2 . 2010 / 2011

= 1005/2011

= 1 - 1/2011

= 2010 / 2011

   1/2 ( 2/1.3 + 2/3.5 +...+ 2 /2009.2011)

= 1/2 ( 1/1 - 1/3 + 1/3 - 1/5 +... + 1/2009 - 1/2011)

= 1/2( 1/1 - 1/2011)

= 1/2 . 2010 / 2011

=1005/2011

\(=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+............+\frac{1}{2009}-\frac{1}{2011}=\frac{1}{1}-\frac{1}{2011}=\frac{2010}{2011}\)

sai rồi top scorer ạ tử trừ mẫu là 2 mà tử là 1 phải nhân 2 lên tử

2A = 2/1.3 +2/3.5 + 2/5.7 + ... + 2/2007.2009 + 2/2009. 2011

2A = 1/1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + ... + 1/ 2007 - 1/2009 + 1/2009 - 1/2011

Gian uoc het ta co: 2A = 1/1 - 1/2011

2A = 2010/2011

A = 2010/2011 X 1/2

A = 1005/2011

**** mình nha 

\(\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+...+\dfrac{1}{2009\cdot2011}\)

\(=\dfrac{1}{2}\cdot\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{2009\cdot2011}\right)\)

\(=\dfrac{1}{2}\cdot\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2009}-\dfrac{1}{2011}\right)\)

\(=\dfrac{1}{2}\cdot\left(1-\dfrac{1}{2011}\right)\)

\(=\dfrac{1}{2}\cdot\dfrac{2010}{2011}=\dfrac{1005}{2011}\)

= 1/2 . (1/1 - 1/3 + 1/3 - 1/5 +... + 1/2009 - 1/2011)

= 1/2 . (1/1 - 1/2011)

= 1/2 . 2010 / 2011

= 1005/2011

= \(\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+....+\frac{2}{2009.2011}\right)\)

= \(\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{2009}-\frac{1}{2011}\right)\)

= \(\frac{1}{2}\left(1-\frac{1}{2011}\right)=\frac{1}{2}.\frac{2010}{2011}=\frac{1005}{2011}\)

\(\frac{1}{1\times3}+\frac{1}{3\times5}+\frac{1}{5\times7}+...+\frac{1}{2009\times2011}\)

= \(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2009}-\frac{1}{2011}\)

= \(1-\frac{1}{2011}\)

= \(\frac{2010}{2011}\)

Đặt A=1/1.3+1/3.5+1/5.7+...+1/2009.2011

2A=2/1.3+2/3.5+2/5.7+...+2/2009.2011

2A=1/1-1/3+1/3-1/5+1/5-1/7+...+1/2009-1/2011

2A=1-1/2011=2011/2011-1/2011=2010/2011

  A=2010/2011.1/2=1005/2011
 

Đặt A=1/1.3+1/3.5+1/5.7+...+1/2009.2011

2A=2/1.3+2/3.5+2/5.7+...+2/2009.2011

2A=1/1-1/3+1/3-1/5+1/5-1/7+...+1/2009-1/2011

2A=1-1/2011=2011/2011-1/2011=2010/2011

A=2010/2011:2

  =1005/2011 

\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2007}-\dfrac{1}{2009}\right)=\dfrac{1}{2}\cdot\dfrac{2008}{2009}=\dfrac{1004}{2009}\)

\(A=\frac{1^2}{1.3}+\frac{2^2}{3.5}+...+\frac{1006^2}{2011.2013}\)

\(\Leftrightarrow4A=\frac{2^2.1^2}{2^2-1}+\frac{2^2.2^2}{4^2-1}+...+\frac{2^2.1006^2}{2012^2-1}\)

\(=1006+\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{2011.2013}\right)\)

\(=1006+\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2011}-\frac{1}{2013}\right)\)

\(=1006+\frac{1}{2}\left(1-\frac{1}{2013}\right)=\frac{2026084}{2013}\)

\(\Rightarrow A=\frac{506521}{2013}\)

1/1.3+1/3.5+1/5.7+...=1/2009.2011

=1/2.(1-1/3+1/3-1/5+1/5-1/7+...+1/2009-1/2011)

=1/2.(1-1/2011)

=1/2.2010/2011

=1005/2011

Gọi tổng trên là A

2A = 2/1.3 + 2/3.5 + 2/5.7 +......+ 2/2009.2011

2A = 1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 +..........+ 1/2009 - 1/2011

2A = 1 - 1/2011

2A = 2010/2011

A = 1005/2011

Vậy................

=1/2(1/1-1/3+1/3-1/5+...-1/2011)

=1/2(1-1/2011)

=1005/2011