Problem: 21+20+2−12−2+2−3+2−4\dfrac{2^{1}+2^{0}+2^{-1}}{2^{-2}+2^{-3}+2^{-4}}2−2+2−3+2−421+20+2−1 equals
Answer Choices:
A. 666
B. 888
C. 312\dfrac{31}{2}231
D. 242424
E. 512512512
Solution:
By inspection, we see that the fraction is of the form
8a+8b+8ca+b+c=8(a+b+c)a+b+c=8\dfrac{8 a+8 b+8 c}{a+b+c}=\dfrac{8(a+b+c)}{a+b+c}=8 a+b+c8a+8b+8c=a+b+c8(a+b+c)=8
Or, compute
2+1+12122+123+124=7214+18+116=72716=8\dfrac{2+1+\dfrac{1}{2}}{\dfrac{1}{2^{2}}+\dfrac{1}{2^{3}}+\dfrac{1}{2^{4}}}=\dfrac{\dfrac{7}{2}}{\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}}=\dfrac{\dfrac{7}{2}}{\dfrac{7}{16}}=8 221+231+2412+1+21=41+81+16127=16727=8