Discussion
\[ \frac{2^{n+4} - 2 \cdot 2^{n}}{2 \cdot 2^{n+3}} + 2^{-3} \ \text{ is equal to:} \]
Answer: D
Factor \( 2^{n} \) out of the fraction.
\[ \frac{2^{n+4} - 2 \cdot 2^{n}}{2 \cdot 2^{n+3}} = \frac{2^{n}(16 - 2)}{2^{n}(16)} = \frac{14}{16} = \frac{7}{8} \]
\[ \frac{7}{8} + 2^{-3} = \frac{7}{8} + \frac{1}{8} = 1 \]
The \( 2^{n} \) cancels entirely, so the answer does not depend on \( n \).
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