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

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

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

simplify

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

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

Solution steps

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

Simplify

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

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

Apply the fraction rule:

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

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

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

6 x^{2} - 3 x - 9

0.0016

2 x + 4 > 12

(5 b - 3) (3 b + 1) = 1

4 a^{2} + 8 a b^{2} + 4 b^{4}

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