What is expand-8/(n^3)[(n(n+1)(2n+1))/4 ] ?

The solution to expand-8/(n^3)[(n(n+1)(2n+1))/4 ] is -4-6/n-2/(n^2)

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expand-8/(n^3)[(n(n+1)(2n+1))/4 ]

expand

- \frac{8}{n ^{3}} [\frac{n ( n + 1 ) ( 2 n + 1 )}{4}]

- 4 - \frac{6}{n} - \frac{2}{n ^{2}}

Solution steps

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

Remove parentheses:

(a) = a

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

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

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

Multiply fractions:

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

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

Cancel

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

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

Expand

2 (2 n^{2} + 3 n + 1) : 4 n^{2} + 6 n + 2

= - \frac{4 n ^{2} + 6 n + 2}{n ^{2}}

Apply the fraction rule:

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

\frac{4 n ^{2} + 6 n + 2}{n ^{2}} = - (\frac{4 n ^{2}}{n ^{2}}) - (\frac{6 n}{n ^{2}}) - (\frac{2}{n ^{2}})

= - (\frac{4 n ^{2}}{n ^{2}}) - (\frac{6 n}{n ^{2}}) - (\frac{2}{n ^{2}})

Remove parentheses:

(a) = a

= - \frac{4 n ^{2}}{n ^{2}} - \frac{6 n}{n ^{2}} - \frac{2}{n ^{2}}

Cancel

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

= - 4 - \frac{6 n}{n ^{2}} - \frac{2}{n ^{2}}

Cancel

\frac{6 n}{n ^{2}} : \frac{6}{n}

= - 4 - \frac{6}{n} - \frac{2}{n ^{2}}

x^{4} + (x^{5} + 2) \frac{d y}{d x}

(- 1 - i) e^{t} (cos (t) + i sin (t))

\frac{( 3 - 2 x )}{9}

sin (\frac{2}{3} (x + 6 π))

\frac{s + e ^{- 2 s}}{s ( s + 1 )}

What is expand-8/(n^3)[(n(n+1)(2n+1))/4 ] ?
The solution to expand-8/(n^3)[(n(n+1)(2n+1))/4 ] is -4-6/n-2/(n^2)