What is ((2^n+2^{n-1}))/((2^{n-1)-2^n)} ?

The solution to ((2^n+2^{n-1}))/((2^{n-1)-2^n)} is -3

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simplify ((2^n+2^{n-1}))/((2^{n-1)-2^n)}

simplify

\frac{( 2 ^{n} + 2 ^{n - 1} )}{( 2 ^{n - 1} - 2 ^{n} )}

- 3

Solution steps

\frac{2 ^{n} + 2 ^{n - 1}}{2 ^{n - 1} - 2 ^{n}}

Factor

2^{n} + 2^{n - 1} : (1 + \frac{1}{2}) \cdot 2^{n}

= \frac{( 1 + \frac{1}{2} ) \cdot 2 ^{n}}{2 ^{n - 1} - 2 ^{n}}

Factor

2^{n - 1} - 2^{n} : (\frac{1}{2} - 1) \cdot 2^{n}

= \frac{( 1 + \frac{1}{2} ) \cdot 2 ^{n}}{( \frac{1}{2} - 1 ) \cdot 2 ^{n}}

Cancel the common factor:

2^{n}

= \frac{1 + \frac{1}{2}}{\frac{1}{2} - 1}

Join

\frac{1}{2} - 1 : - \frac{1}{2}

= \frac{1 + \frac{1}{2}}{- \frac{1}{2}}

Join

1 + \frac{1}{2} : \frac{3}{2}

= \frac{\frac{3}{2}}{- \frac{1}{2}}

Apply the fraction rule:

\frac{a}{- b} = - \frac{a}{b}

= - \frac{\frac{3}{2}}{\frac{1}{2}}

Divide fractions:

\frac{\frac{a}{b}}{\frac{c}{d}} = \frac{a \cdot d}{b \cdot c}

= - \frac{3 \cdot 2}{2 \cdot 1}

Refine

= - \frac{3 \cdot 2}{2}

Cancel the common factor:

2

= - 3

(\frac{p ^{2} - 3}{2 \cdot q ^{2}})^{3}

(3 x^{4} \cdot y^{- 5})^{3}

(sec (x) - tan (x)) 2

(- 5) (2 a - 3 b + 5 c) - 4 (2 a + 6 - 2 c)

(2 x - 4)^{3} - (3 x + 5)^{3}

What is ((2^n+2^{n-1}))/((2^{n-1)-2^n)} ?
The solution to ((2^n+2^{n-1}))/((2^{n-1)-2^n)} is -3