Question

Va rog ajutati-ma!
Sa se calculeze: $\frac{2}{1}$+$\frac{4}{3}$+$\frac{6}{5}$+$\frac{8}{7}$+...+$\frac{2004}{2003}$-(1+$\frac{1}{3}$+$\frac{1}{5}$+$\frac{1}{7}$+...+$\frac{1}{2003}$).

Answer 1

$\displaystyle \frac{2}{1}+ \frac{4}{3}+ \frac{6}{5}+ \frac{8}{7}+...+ \frac{2004}{2003}- \left( 1+ \frac{1}{3}+ \frac{1}{5}+ \frac{1}{7}+...+ \frac{1}{2003}\right)= \\ \\ = \frac{2}{1}+ \frac{4}{3}+ \frac{6}{5}+ \frac{8}{7}+...+ \frac{2004}{2003}- 1- \frac{1}{3}- \frac{1}{5}- \frac{1}{7}-...- \frac{1}{2003}\right)=$

$\displaystyle = \left( \frac{2}{1}-1 \right) + \left( \frac{4}{3} - \frac{1}{3} \right)+\left( \frac{6}{5} - \frac{1}{5} \right)+\left( \frac{8}{7} - \frac{1}{7} \right)+...+ \\ \\ +\left( \frac{2004}{2003} - \frac{1}{2003} \right)= \\ \\ = 1+\frac{3}{3}+ \frac{5}{5}+ \frac{7}{7}+...+ \frac{2003}{2003}= \\ \\ =1+1+1+1+...+1= \\ \\ =1002.$