Question
Simplify: $\frac{16\times(2)^{\text{n+1}}-4\times2^{\text{n}}}{16\times(2)^{\text{n}+2}-2\times(2)^{\text{n}+2}}$
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Answer
$\frac{16\times(2)^{\text{n+1}}-4\times2^{\text{n}}}{16\times(2)^{\text{n}+2}-2\times(2)^{\text{n}+2}}$
$=\frac{2^4\times2^{(\text{n}+1)}-2^2\times2^{\text{n}}}{2^4\times2^{(\text{n}+2)}-2^{\text{n}+1}\times2^2}$
$=\frac{2^2\times2^{(\text{n}+3-2\text{n})}}{2^2\times2^{(\text{n}+4-2\text{n}+1)}}$
$=\frac{2^{\text{n}}\times2^3-2^{\text{n}}}{2^{\text{n}}(2^{4}-1)}$
$=\frac{2^{\text{n}}(2^3-1)}{2^{\text{n}}(2^4-1)}$
$=\frac{8-1}{16-2}$
$=\frac{7}{14}$
$=\frac{1}{2}$
$=\frac{2^4\times2^{(\text{n}+1)}-2^2\times2^{\text{n}}}{2^4\times2^{(\text{n}+2)}-2^{\text{n}+1}\times2^2}$
$=\frac{2^2\times2^{(\text{n}+3-2\text{n})}}{2^2\times2^{(\text{n}+4-2\text{n}+1)}}$
$=\frac{2^{\text{n}}\times2^3-2^{\text{n}}}{2^{\text{n}}(2^{4}-1)}$
$=\frac{2^{\text{n}}(2^3-1)}{2^{\text{n}}(2^4-1)}$
$=\frac{8-1}{16-2}$
$=\frac{7}{14}$
$=\frac{1}{2}$
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