[tex](1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99})-(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100})=\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}[/tex]
\[\begin{align}
& =\left( 1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100} \right)-2.\left( \frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100} \right) \\
& =\left( 1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100} \right)-(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}) \\
& =\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100} \\
\end{align}\]