$$\frac{{{2^{n + 4}} - 2\left( {{2^n}} \right)}}{{2\left( {{2^{n + 3}}} \right)}}$$    when simplified is = ?

A. $${{\text{2}}^{n + 1}} - \frac{1}{8}$$

B. -2⁽ⁿ⁺¹⁾

C. 1 - 2ⁿ

D. $$\frac{7}{8}$$

Answer: Option D

Solution(By Examveda Team)

$$\eqalign{ & \frac{{{2^{n + 4}} - 2\left( {{2^n}} \right)}}{{2\left( {{2^{n + 3}}} \right)}}{\text{ }} \cr & = \frac{{{2^{n + 4}} - {2^{n + 1}}}}{{{2^{n + 4}}}}{\text{ }} \cr & = \frac{{{2^{n + 4}}}}{{{2^{n + 4}}}} - \frac{{{2^{n + 1}}}}{{{2^{n + 4}}}}{\text{ }} \cr & = 1 - {2^{(n + 1) - \left( {n + 4} \right)}} \cr & {\text{ = 1}} - {2^{ - 3}} \cr & {\text{ = 1}} - \frac{1}{8}{\text{ }} \cr & {\text{ = }}\frac{7}{8} \cr} $$