What is expand ((n(n+1)(2n+1))/6)(2/n)^3 ?

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

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expand ((n(n+1)(2n+1))/6)(2/n)^3

expand

(\frac{n ( n + 1 ) ( 2 n + 1 )}{6}) (\frac{2}{n})^{3}

\frac{8}{3} + \frac{4}{n} + \frac{4}{3 n ^{2}}

Solution steps

(\frac{n ( n + 1 ) ( 2 n + 1 )}{6}) (\frac{2}{n})^{3}

Remove parentheses:

(a) = a

= \frac{n ( n + 1 ) ( 2 n + 1 )}{6} (\frac{2}{n})^{3}

\frac{n ( n + 1 ) ( 2 n + 1 )}{6} = \frac{2 n ^{3} + 3 n ^{2} + n}{6}

= (\frac{2}{n})^{3} \frac{2 n ^{3} + 3 n ^{2} + n}{6}

(\frac{2}{n})^{3} = \frac{2 ^{3}}{n ^{3}}

= \frac{2 ^{3}}{n ^{3}} \cdot \frac{2 n ^{3} + 3 n ^{2} + n}{6}

Multiply fractions:

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

= \frac{( 2 n ^{3} + 3 n ^{2} + n ) \cdot 2 ^{3}}{6 n ^{3}}

Factor

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

= \frac{8 n ( 2 n ^{2} + 3 n + 1 )}{6 n ^{3}}

Cancel

\frac{8 n ( 2 n ^{2} + 3 n + 1 )}{6 n ^{3}} : \frac{4 ( 2 n ^{2} + 3 n + 1 )}{3 n ^{2}}

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

Expand

4 (2 n^{2} + 3 n + 1) : 8 n^{2} + 12 n + 4

= \frac{8 n ^{2} + 12 n + 4}{3 n ^{2}}

Apply the fraction rule:

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

\frac{8 n ^{2} + 12 n + 4}{3 n ^{2}} = \frac{8 n ^{2}}{3 n ^{2}} + \frac{12 n}{3 n ^{2}} + \frac{4}{3 n ^{2}}

= \frac{8 n ^{2}}{3 n ^{2}} + \frac{12 n}{3 n ^{2}} + \frac{4}{3 n ^{2}}

Cancel

\frac{8 n ^{2}}{3 n ^{2}} : \frac{8}{3}

= \frac{8}{3} + \frac{12 n}{3 n ^{2}} + \frac{4}{3 n ^{2}}

Cancel

\frac{12 n}{3 n ^{2}} : \frac{4}{n}

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

\frac{2 ( - 2 e ^{- x} cos ( 2 x ) - e ^{- x} sin ( 2 x ) )}{5}

(x^{2} y - x^{2} + y - 1) d x + (x y + 2 x - 3 y - 6) d y

\frac{s + 3}{( s - 2 ) ( s + 1 )}

\frac{\partial}{\partial x} ((f (x, y)) (4 y^{3} + 2 y))

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