$解:2×(1+\frac{1}{2})×(1+\frac{1}{2^2})×(1+\frac{1}{2^4})×(1+\frac{1}{2^8})+\frac{1}{2^{14}}$
$=4×(1-\frac{1}{2})×(1+\frac{1}{2})×(1+\frac{1}{2^2})×(1+\frac{1}{2^4})×(1+ \frac{1}{2^8})+\frac{1}{2^{14}}$
$=4×(1\frac{1}{22}) ×(1+\frac{1}{2^2})×(1+\frac{1}{2^4})×(1+ \frac{1}{2^8})+\frac{1}{2^{14}}$
$=4×(1-\frac{1}{2^4}) ×(1+\frac{1}{2^4})×(1+ \frac{1}{2^8})+\frac{1}{2^{14}}\ $
$=4×(1-\frac{1}{2^8})×(1+ \frac{1}{2^8})+\frac{1}{2^{14}}\ $
$=4×(1-\frac{1}{2^{16}}+\frac{1}{2^{14}} =4-\frac{1}{2^{14}}+\frac{1}{2^{14}}$
$=4$