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1. $\frac{2^{n + 4} - 2 \times 2^{n}}{2 \times 2^{(n + 3)}} + 2^{- 3}$ is equal to = ?

A. 2⁽ⁿ⁺¹⁾

B. $(\frac{9}{8} - 2^{n})$

C. $(- 2^{n + 1} + \frac{1}{8})$

D. 1

Solution:

\begin{aligned} \frac{2^{n + 4} - 2 \times 2^{n}}{2 \times 2^{(n + 3)}} + 2^{- 3} \\ = \frac{2^{n + 4} - 2^{n + 1}}{2^{(n + 4)}} + \frac{1}{2^{3}} \\ = \frac{2^{n + 1} (2^{3} - 1)}{2^{(n + 4)}} + \frac{1}{2^{3}} \\ = \frac{2^{n + 1} \times 7}{2^{n + 1} \times 2^{3}} + \frac{1}{2^{3}} \\ = (\frac{7}{8} + \frac{1}{8}) \\ = \frac{8}{8} \\ = 1 \end{aligned}

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