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

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

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

simplify

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

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

Solution steps

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

Apply exponent rule:

(a^{b})^{c} = a^{b c}, a \geq 0

(2^{n})^{n - 1} = 2^{n (n - 1)}

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

Simplify

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

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

\frac{7 b - 14 b ^{2}}{42 b ^{2} - 21 b}

(- 2.3551 \cdot 1 0^{4}) \cdot \frac{1}{8.314} \cdot \frac{1}{285}

4 x + 1 = 0

4 b^{4} - 49 b^{2}

\frac{9}{7} \cdot 14

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