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Question

$$\displaystyle \frac{16\times 2^{n+1}-4\times 2^{n}}{16\times 2^{n+2}-2\times 2^{n+2}}$$ equals


A
14
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B
12
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C
14
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D
12
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Solution

The correct option is D $$\displaystyle \frac{1}{2}$$
$$=\cfrac{ 16\times 2^{n+1}-4\times 2^n }{  16\times 2^{n+2}- 2\times 2^{n+2} }$$
We know, $$a^{m+l}=a^m \times a^l$$
Thus, $$2^{n+1}=2^n \times 2^1$$ 
        
         $$2^{n+2}=2^n \times 2^2$$
$$=\cfrac{ 16\times 2^n \times 2-4\times 2^n }{  16\times 2^n \times 2^2- 2\times 2^n \times 2^2 }$$
$$=\cfrac{2^n(16\times 2-4)}{2^n(116\times 4-2\times 4)} \\ =\cfrac{32-4}{64-8} =\cfrac{28}{56} \\ =\cfrac{1}{2}$$

Mathematics

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