Question
Prove that:$\frac{2^\text{n}+2^{\text{n-1}}}{2^{\text{n}+1}-2^\text{n}}=\frac{3}{2}$
✓
Answer
We have,$\frac{2^\text{n}+2^{\text{n-1}}}{2^{\text{n}+1}-2^\text{n}}=\frac{2^\text{n}+2^\text{n}\times2^{-1}}{2^\text{n}\times\text{2}^1-2^\text{n}}$
$=\frac{2^\text{n}[1+2^{-1}]}{2^\text{n}[2-1]}$
$=\frac{1+\frac{1}{2}}{1}$
$=1+\frac{1}{2}$
$=\frac{3}{2}$
$\Rightarrow\frac{2^\text{n}+2^{\text{n}-1}}{2^{\text{n}+1}-2^\text{n}}=\frac{3}{2}$
$=\frac{2^\text{n}[1+2^{-1}]}{2^\text{n}[2-1]}$
$=\frac{1+\frac{1}{2}}{1}$
$=1+\frac{1}{2}$
$=\frac{3}{2}$
$\Rightarrow\frac{2^\text{n}+2^{\text{n}-1}}{2^{\text{n}+1}-2^\text{n}}=\frac{3}{2}$
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