解：$ ②$原式$=(1-\frac {1}{14})+((1-\frac {1}{28})+(1-\frac {1}{56})+(1-\frac {1}{112})$
$ =4-\frac {1}{7}(\frac {1}{2}+\frac {1}{2^2}+\frac {1}{2^3}+\frac {1}{2^4}) $
$=4-\frac {1}{7}×(1-\frac {1}{2^4})$
$=4-\frac {1}{7}×\frac {15}{16}$
$=\frac {433}{112} $
