Question
Simplify: $\frac{16\times2^{\text{n}+1}-8\times2^{\text{n}}}{16\times2^{\text{n}+2}-4\times2^{\text{n}+1}}$
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Answer
We have, $\frac{10\times2^{\text{n}+1}-2^3\times2^\text{n}}{10\times2^{\text{n}+2}-4\times2^{\text{n}+1}}$
$\Rightarrow \frac{2^4\times2^{\text{n}+1}-2^3\times2^\text{n}}{2^4\times2^{\text{n}+2}-2^2\times2^{\text{n}+1}}$
$\Rightarrow \frac{2^3(2^{\text{n}+2}-2^\text{n})}{2^3\times(2^{\text{n}+3}-2^\text{n})}$
$\Rightarrow \frac{2^\text{n}\times2^2-2^\text{n}}{2^\text{n}\times2^3-2^\text{n}}$
$\Rightarrow \frac{2^\text{n}(2^2-1)}{2^\text{n}(2^3-1)}=\frac{4-1}{8-1}$
$=\frac{3}{7}$
$\Rightarrow \frac{2^4\times2^{\text{n}+1}-2^3\times2^\text{n}}{2^4\times2^{\text{n}+2}-2^2\times2^{\text{n}+1}}$
$\Rightarrow \frac{2^3(2^{\text{n}+2}-2^\text{n})}{2^3\times(2^{\text{n}+3}-2^\text{n})}$
$\Rightarrow \frac{2^\text{n}\times2^2-2^\text{n}}{2^\text{n}\times2^3-2^\text{n}}$
$\Rightarrow \frac{2^\text{n}(2^2-1)}{2^\text{n}(2^3-1)}=\frac{4-1}{8-1}$
$=\frac{3}{7}$
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