Compute $\displaystyle \frac{2+4-8+16+32-64}{4+8-16+32+64-128}$.
Factoring the numerator and denominator, we have:

$\displaystyle \frac{2+4-8+16+32-64}{4+8-16+32+64-128}=\frac{2(1+2-4+8+16-32)}{4(1+2-4+8+16-32)}=\frac{2}{4}=\boxed{\frac{1}{2}}$.