1951 AHSME Problems/Problem 24

Problem

$\frac{2^{n+4}-2(2^{n})}{2(2^{n+3})}$ when simplified is:

$\textbf{(A)}\ 2^{n+1}-\frac{1}{8}\qquad\textbf{(B)}\ -2^{n+1}\qquad\textbf{(C)}\ 1-2^{n}\qquad\textbf{(D)}\ \frac{7}{8}\qquad\textbf{(E)}\ \frac{7}{4}$

Solution

We have $2(2^n)=2^{n+1}$, and $2(2^{n+3})=2^{n+4}$. Thus, $\frac{2^{n+4}-2(2^{n})}{2(2^{n+3})}=\dfrac{2^{n+4}-2^{n+1}}{2^{n+4}}$. Factoring out a $2^{n+1}$ in the numerator, we get $\dfrac{2^{n+1}(2^3-1)}{2^{n+4}}=\dfrac{8-1}{2^3}=\boxed{\textbf{(D)}\ \frac{7}{8}}$.

See Also

 1951 AHSC (Problems • Answer Key • Resources) Preceded byProblem 23 Followed byProblem 25 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 All AHSME Problems and Solutions

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