What is expand (4(-1)^{n+1}n-8n+(-1)^n-1)/4 ?

The solution to expand (4(-1)^{n+1}n-8n+(-1)^n-1)/4 is (-1)^{n+1}n-2n+((-1)^n)/4-1/4

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expand (4(-1)^{n+1}n-8n+(-1)^n-1)/4

expand

\frac{4 ( - 1 ) ^{n + 1} n - 8 n + ( - 1 ) ^{n} - 1}{4}

(- 1)^{n + 1} n - 2 n + \frac{( - 1 ) ^{n}}{4} - \frac{1}{4}

Solution steps

\frac{4 ( - 1 ) ^{n + 1} n - 8 n + ( - 1 ) ^{n} - 1}{4}

Apply the fraction rule:

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

\frac{4 ( - 1 ) ^{n + 1} n - 8 n + ( - 1 ) ^{n} - 1}{4} = \frac{4 ( - 1 ) ^{n + 1} n}{4} - \frac{8 n}{4} + \frac{( - 1 ) ^{n}}{4} - \frac{1}{4}

= \frac{4 ( - 1 ) ^{n + 1} n}{4} - \frac{8 n}{4} + \frac{( - 1 ) ^{n}}{4} - \frac{1}{4}

Divide the numbers:

\frac{4}{4} = 1

= (- 1)^{n + 1} n - \frac{8 n}{4} + \frac{( - 1 ) ^{n}}{4} - \frac{1}{4}

Divide the numbers:

\frac{8}{4} = 2

= (- 1)^{n + 1} n - 2 n + \frac{( - 1 ) ^{n}}{4} - \frac{1}{4}

(2 x + 4 y) + (2 x - y) \frac{d y}{d x}

\frac{2 ( 8 2 - 2 )}{5}

(- 1, - 1) y (3, - 2)

\frac{2 ( 1 + 10 )}{3}

(x + y i) (x - y i)^{2}

What is expand (4(-1)^{n+1}n-8n+(-1)^n-1)/4 ?
The solution to expand (4(-1)^{n+1}n-8n+(-1)^n-1)/4 is (-1)^{n+1}n-2n+((-1)^n)/4-1/4