| If \[\frac{{{2}^{n+4}}-2\cdot {{2}^{n}}}{2\cdot {{2}^{n+3}}}+{{2}^{-\,\,3}}=x,\]then the value of x is |
A) \[-\,\,{{2}^{n+1}}+\frac{1}{8}\]
B) \[1\]
C) \[{{2}^{n+1}}\]
D) \[\frac{n}{8}-{{2}^{n}}\]
Correct Answer: B
Solution :
| \[\frac{{{2}^{n+4}}-2\cdot {{2}^{n}}}{2\cdot {{2}^{n+3}}}+{{2}^{-3}}=x\] |
| \[\Rightarrow \] \[x=\frac{{{2}^{n+4}}-{{2}^{n+1}}}{{{2}^{n+4}}}+{{2}^{\,\,-\,\,3}}\] |
| \[=\frac{{{2}^{n+1}}({{2}^{3}}-1)}{{{2}^{n+4}}}+\frac{1}{{{2}^{3}}}\] |
| \[=\frac{8-1}{{{2}^{3}}}+\frac{1}{{{2}^{3}}}=\frac{7}{8}+\frac{1}{8}=1\] |
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