解:原式$= 2×\frac 12×(1+\frac 12)(1+\frac 14)×(1+\frac 1{16})$
$ = 2×(1-\frac 12)(1+\frac 12)(1+\frac 14)(1+\frac 1{16})$
$ = 2×(1-\frac 14)(1+\frac 14)(1+\frac 1{16})$
$ = 2×(1-\frac 1{16})(1+\frac 1{16})$
$ = 2×\frac 1{16²}$
$ = 2×\frac {255}{256}$
$ = \frac {255}{128}$