Question

tính

1/1.3+1/3.5+15.7+...+1/2009.2011

Answer 1

\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2009.2011}\)

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

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

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

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

Answer 2

\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2009}-\frac{1}{2011}\right)=\frac{1}{2}.\left(1-\frac{1}{2011}\right)=\frac{1}{2}.\frac{2010}{2011}=\frac{1005}{2011}\)