A) \[\frac{1}{4}\]
B) \[\frac{1}{8}\]
C) \[\frac{1}{2}\]
D) \[\frac{1}{16}\]
Correct Answer: C
Solution :
Expression \[=\frac{16\times {{2}^{n+1}}-4\times {{2}^{n}}}{16\times {{2}^{n+2}}-2\times {{2}^{n+2}}}\] \[=\frac{{{2}^{4}}\times {{2}^{n+1}}-{{2}^{2}}\times {{2}^{n}}}{{{2}^{4}}\times {{2}^{n+2}}-2\times {{2}^{n+2}}}\] \[=\frac{{{2}^{n+5}}-{{2}^{n+2}}}{{{2}^{n+6}}-{{2}^{n+3}}}=\frac{{{2}^{n+5}}-{{2}^{n+2}}}{{{2.2}^{n+5}}-{{2.2}^{n+2}}}\] \[=\frac{{{2}^{n+5}}-{{2}^{n+2}}}{2({{2}^{n+5}}-{{2}^{n+2}})}=\,\frac{1}{2}\]
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