$(1+\frac{1}{2})×(1-\frac{1}{2})×(1+\frac{1}{3})×(1-\frac{1}{3})×...× (1+\frac{1}{99})×(1-\frac{1}{99})$
$=\frac{3}{2}×\frac{1}{2}×\frac{4}{3}×\frac{2}{3}×...×\frac{100}{99}×\frac{98}{99}$
$=(\frac{3}{2}×\frac{2}{3})×(\frac{4}{3}×\frac{3}{4})×...×(\frac{99}{98}×\frac{98}{99})×(\frac{1}{2}×\frac{100}{99})$
$=\frac{1}{2}×\frac{100}{99}$
$=\frac{50}{99}$