What is expand (9*2)/n-((n(n+1)(2n+1))/6)(2/n)^3 ?

The solution to expand (9*2)/n-((n(n+1)(2n+1))/6)(2/n)^3 is -8/3+(14)/n-4/(3n^2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

expand (9*2)/n-((n(n+1)(2n+1))/6)(2/n)^3

expand

\frac{9 \cdot 2}{n} - (\frac{n ( n + 1 ) ( 2 n + 1 )}{6}) (\frac{2}{n})^{3}

- \frac{8}{3} + \frac{14}{n} - \frac{4}{3 n ^{2}}

Solution steps

\frac{9 \cdot 2}{n} - (\frac{n ( n + 1 ) ( 2 n + 1 )}{6}) (\frac{2}{n})^{3}

Remove parentheses:

(a) = a

= \frac{9 \cdot 2}{n} - \frac{n ( n + 1 ) ( 2 n + 1 )}{6} (\frac{2}{n})^{3}

\frac{9 \cdot 2}{n} = \frac{18}{n}

\frac{n ( n + 1 ) ( 2 n + 1 )}{6} (\frac{2}{n})^{3} = \frac{8 n ^{2} + 12 n + 4}{3 n ^{2}}

= \frac{18}{n} - \frac{8 n ^{2} + 12 n + 4}{3 n ^{2}}

Least Common Multiplier of

n, 3 n^{2} : 3 n^{2}

Adjust Fractions based on the LCM

= \frac{54 n}{3 n ^{2}} - \frac{8 n ^{2} + 12 n + 4}{3 n ^{2}}

Since the denominators are equal, combine the fractions:

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

= \frac{54 n - ( 8 n ^{2} + 12 n + 4 )}{3 n ^{2}}

Expand

54 n - (8 n^{2} + 12 n + 4) : - 8 n^{2} + 42 n - 4

= \frac{- 8 n ^{2} + 42 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} + 42 n - 4}{3 n ^{2}} = - \frac{8 n ^{2}}{3 n ^{2}} + \frac{42 n}{3 n ^{2}} - \frac{4}{3 n ^{2}}

= - \frac{8 n ^{2}}{3 n ^{2}} + \frac{42 n}{3 n ^{2}} - \frac{4}{3 n ^{2}}

Cancel

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

= - \frac{8}{3} + \frac{42 n}{3 n ^{2}} - \frac{4}{3 n ^{2}}

Cancel

\frac{42 n}{3 n ^{2}} : \frac{14}{n}

= - \frac{8}{3} + \frac{14}{n} - \frac{4}{3 n ^{2}}

lo g_{10} (10) (x) - lo g_{10} (10) (x)

\frac{x + 2}{( x + 1 ) ^{2} ( x + 3 )}

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

\frac{d}{d x} (- a (x) e^{b (x - 190)}) - 15

(a + b) (a - b) (a^{2} - b) + b

What is expand (9*2)/n-((n(n+1)(2n+1))/6)(2/n)^3 ?
The solution to expand (9*2)/n-((n(n+1)(2n+1))/6)(2/n)^3 is -8/3+(14)/n-4/(3n^2)