What is expand-8/(n^3)((n(n+1))/2)^2 ?

The solution to expand-8/(n^3)((n(n+1))/2)^2 is -2n-4-2/n

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expand-8/(n^3)((n(n+1))/2)^2

expand

- \frac{8}{n ^{3}} (\frac{n ( n + 1 )}{2})^{2}

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

Solution steps

- \frac{8}{n ^{3}} (\frac{n ( n + 1 )}{2})^{2}

(\frac{n ( n + 1 )}{2})^{2} = \frac{n ^{4} + 2 n ^{3} + n ^{2}}{2 ^{2}}

= - \frac{8}{n ^{3}} \cdot \frac{n ^{4} + 2 n ^{3} + n ^{2}}{2 ^{2}}

Multiply fractions:

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

= - \frac{8 ( n ^{4} + 2 n ^{3} + n ^{2} )}{n ^{3} \cdot 2 ^{2}}

Cancel

\frac{8 ( n ^{4} + 2 n ^{3} + n ^{2} )}{n ^{3} \cdot 2 ^{2}} : \frac{2 ( n ^{2} + 2 n + 1 )}{n}

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

Expand

2 (n^{2} + 2 n + 1) : 2 n^{2} + 4 n + 2

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

Apply the fraction rule:

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

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

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

Remove parentheses:

(a) = a

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

Cancel

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

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

Cancel

\frac{4 n}{n} : 4

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

(2 x^{2} y + \frac{1}{x y ^{2}})^{7}

\frac{1}{( 1 - x ^{4} )} \cdot \frac{1}{( 1 - x ^{5} )}

sin^{2} (u) (v^{2} + 1) + cos^{2} (u) (v^{2} + 1) + v^{2}

(\times + \frac{1}{2}) (x - \frac{1}{2})

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

What is expand-8/(n^3)((n(n+1))/2)^2 ?
The solution to expand-8/(n^3)((n(n+1))/2)^2 is -2n-4-2/n