What is expand (n+1)-((n+1)^2)/(4n) ?

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

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expand (n+1)-((n+1)^2)/(4n)

expand

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

n + 1 - \frac{n}{4} - \frac{1}{2} - \frac{1}{4 n}

Solution steps

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

Remove parentheses:

(a) = a

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

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

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

Apply the fraction rule:

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

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

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

Remove parentheses:

(a) = a

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

Cancel

\frac{n ^{2}}{4 n} : \frac{n}{4}

= n + 1 - \frac{n}{4} - \frac{2 n}{4 n} - \frac{1}{4 n}

Cancel

\frac{2 n}{4 n} : \frac{1}{2}

= n + 1 - \frac{n}{4} - \frac{1}{2} - \frac{1}{4 n}

x (2 x - 7 = 1)

(x - 3) e^{\frac{1}{x}}

A (- 3, 2) B (7, - 3)

5 + 5 p + p + 5

(3 x - 2) \cdot e^{x}

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