$$\frac{{{2^{n + 4}} - 2 \times {2^n}}}{{2 \times {2^{\left( {n + 3} \right)}}}} + {2^{ - 3}}$$ is equal to = ?
A. 2(n+1)
B. $$\left( {\frac{9}{8} - {2^n}} \right)$$
C. $$\left( { - {2^{n + 1}} + \frac{1}{8}} \right)$$
D. 1
Answer: Option D
Solution (By Examveda Team)
$$\eqalign{ & \frac{{{2^{n + 4}} - 2 \times {2^n}}}{{2 \times {2^{\left( {n + 3} \right)}}}} + {2^{ - 3}} \cr & = \frac{{{2^{n + 4}} - {2^{n + 1}}}}{{{2^{\left( {n + 4} \right)}}}} + \frac{1}{{{2^3}}} \cr & = \frac{{{2^{n + 1}}\left( {{2^3} - 1} \right)}}{{{2^{\left( {n + 4} \right)}}}} + \frac{1}{{{2^3}}} \cr & = \frac{{{2^{n + 1}} \times 7}}{{{2^{n + 1}} \times {2^3}}} + \frac{1}{{{2^3}}} \cr & = \left( {\frac{7}{8} + \frac{1}{8}} \right) \cr & = \frac{8}{8} \cr & = 1 \cr} $$This Question Belongs to Arithmetic Ability >> Surds And Indices
Join The Discussion