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

$\displaystyle \frac{2+4-8+16+32-64}{4+8-16+32+64-128}=\frac{2(1+2-4+8+16-32)}{4(1+2-4+8+16-32)}=\frac{2}{4}=\boxed{\frac{1}{2}}$.
\frac{1}{2}
Level 2
Algebra
Problem: Compute $\displaystyle \frac{2+4-8+16+32-64}{4+8-16+32+64-128}$.
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To solve the problem, first simplify the numerator and denominator separately. 

For the numerator, calculate each term: 
2 + 4 = 6, 
6 - 8 = -2, 
-2 + 16 = 14, 
14 + 32 = 46, 
46 - 64 = -18. 
Thus, the numerator simplifies to -18.

For the denominator, calculate each term: 
4 + 8 = 12, 
12 - 16 = -4, 
-4 + 32 = 28, 
28 + 64 = 92, 
92 - 128 = -36. 
Thus, the denominator simplifies to -36.

Now, compute the fraction: 
\frac{-18}{-36}. 
Since both the numerator and denominator are negative, the fraction simplifies to \frac{18}{36}.

Next, simplify \frac{18}{36} by dividing both the numerator and denominator by their greatest common divisor, which is 18: 
\frac{18 \div 18}{36 \div 18} = \frac{1}{2}.

Therefore, the final answer is \frac{1}{2}.<|END_OF_LATENT|> The final answer is $\frac{1}{2}$.