Bài giải :

$\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+...+2009}$

= $\frac{\text{1}}{\text{(1 + 2) x 2 : 2}}\text{+}\frac{\text{1}}{\text{(1 + 3) x 3 : 2}}\text{+}\frac{\text{1}}{\text{(1+4) x 4 : 2}}\text{+}...\text{+}\frac{\text{1}}{\text{(1 + 2009) x 2009 : 2}}$

 = $\frac{\text{2}}{\text{2x3}}\text{+}\frac{\text{2}}{\text{3x4}}\text{+}\frac{\text{2}}{\text{4x5}}\text{+}...\text{+}\frac{\text{2}}{\text{2009x2010}}$

 = $\text{2 x (}\frac{\text{1}}{\text{2 x 3}}\text{+}\frac{\text{1}}{\text{3 x 4}}\text{+}\frac{\text{1}}{\text{4 x 5}}\text{+}...\text{+}\frac{\text{1}}{\text{2009 x 2010}}\text{)}$

=$\text{2 x (}\frac{\text{1}}{\text{2}}\text{-}\frac{\text{1}}{\text{3}}\text{+}\frac{\text{1}}{\text{3}}\text{-}\frac{\text{1}}{\text{4}}\text{+}\frac{\text{1}}{\text{4}}\text{-}\frac{\text{1}}{\text{5}}\text{+}...\text{+}\frac{\text{1}}{\text{2009}}\text{-}\frac{\text{1}}{\text{2010}}\text{)}$

=$\text{2 x (}\frac{\text{1}}{\text{2}}\text{-}\frac{\text{1}}{\text{2010}}\text{)}$ = $\text{2 x}\,\,\frac{\text{1004}}{\text{2010}}$ = $\frac{\text{1004}}{\text{1005}}$