解:原式$=\frac {1×3+1}{1×3}×\frac {2×4+1}{2×4}×\frac {3×5+1}{3×5}×...× \frac {98×100+1}{98×100}$
$=\frac {22}{1×3}×\frac {32}{2×4}×\frac {4^2}{3×5}×...×\frac {992}{98×100}$
$=2 \frac 23×\frac 32×\frac 34×\frac 43×\frac 45×...×\frac {99}{98}×\frac {99}{100}=\frac 21×\frac {99}{100}$
$=\frac {99}{50}.$