解$:(4)$原式$=(1-\frac {1}{2})×(1+\frac {1}{2})×(1-\frac {1}{3})×(1+\frac {1}{3})×···$
$×(1-\frac {1}{100})×(1+\frac {1}{100})$
$=\frac {1}{2}×\frac {3}{2}×\frac {2}{3}×\frac {4}{3}×\frac {3}{4}×···×\frac {99}{100}×\frac {101}{100}$
$=\frac {101}{200}.$