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

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

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expand (-2^{n-1}n+2^nn-2^n)/(2^{2n-1)}

expand

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

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

Solution steps

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

Factor

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

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

Cancel

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

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

Distribute parentheses

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

Apply minus-plus rules

+ (- a) = - a

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

Simplify

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

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

(x^{2} - 7)^{4} (2 x) d x

\frac{8 ( 16 k ^{2} + 16 kn )}{3 n ^{3}}

(4 (\frac{d I}{d t}) + 12 I) (e^{2 t})

(- 1 + 3 \frac{d y}{d x}) (x - 3)

- ln (sin (x)) cos (x) + sin (x) (- x - cot (x))

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