Question

1 / 2001x2003 + 1/2003x2005 + 1/2005x2007 +...+ 1/2011x2013

Answer 1

\(\dfrac{1}{2001.2003}+\dfrac{1}{2003.2005}+...+\dfrac{1}{2011.2013}\)

\(=\dfrac{1}{2}\left(\dfrac{2}{2001.2003}+\dfrac{1}{2003.2005}+...+\dfrac{1}{2011.2013}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{2001}-\dfrac{1}{2003}+\dfrac{1}{2003}-\dfrac{1}{2005}+...+\dfrac{1}{2011}-\dfrac{1}{2013}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{2001}-\dfrac{1}{2013}\right)=\dfrac{2}{1342671}\)

Answer 2

=1/2.(2/2001.2003+2/2003.2005+.....+2/2011.2013)

=1/2(1/2001-1/2003+1/2003-1/2005+....+1/2011-1/2013)

=1/2(1/2001-1/2013)

=2/1342671