What is expand ((n+1)(6n^4+9n^3+n^2-n))/(30) ?

The solution to expand ((n+1)(6n^4+9n^3+n^2-n))/(30) is (n^5)/5+(n^4)/2+(n^3)/3-n/(30)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

expand ((n+1)(6n^4+9n^3+n^2-n))/(30)

expand

\frac{( n + 1 ) ( 6 n ^{4} + 9 n ^{3} + n ^{2} - n )}{30}

\frac{n ^{5}}{5} + \frac{n ^{4}}{2} + \frac{n ^{3}}{3} - \frac{n}{30}

Solution steps

\frac{( n + 1 ) ( 6 n ^{4} + 9 n ^{3} + n ^{2} - n )}{30}

Expand

(n + 1) (6 n^{4} + 9 n^{3} + n^{2} - n) : 6 n^{5} + 15 n^{4} + 10 n^{3} - n

= \frac{6 n ^{5} + 15 n ^{4} + 10 n ^{3} - n}{30}

Apply the fraction rule:

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

\frac{6 n ^{5} + 15 n ^{4} + 10 n ^{3} - n}{30} = \frac{6 n ^{5}}{30} + \frac{15 n ^{4}}{30} + \frac{10 n ^{3}}{30} - \frac{n}{30}

= \frac{6 n ^{5}}{30} + \frac{15 n ^{4}}{30} + \frac{10 n ^{3}}{30} - \frac{n}{30}

Cancel

\frac{6 n ^{5}}{30} : \frac{n ^{5}}{5}

= \frac{n ^{5}}{5} + \frac{15 n ^{4}}{30} + \frac{10 n ^{3}}{30} - \frac{n}{30}

Cancel

\frac{15 n ^{4}}{30} : \frac{n ^{4}}{2}

= \frac{n ^{5}}{5} + \frac{n ^{4}}{2} + \frac{10 n ^{3}}{30} - \frac{n}{30}

Cancel

\frac{10 n ^{3}}{30} : \frac{n ^{3}}{3}

= \frac{n ^{5}}{5} + \frac{n ^{4}}{2} + \frac{n ^{3}}{3} - \frac{n}{30}

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

\frac{1}{1 - \frac{v ^{2}}{c ^{2}}}

\frac{1}{x ( 1 + t x )}

x y d x - (2 x + 4 y) d y

\frac{2 x}{( z + x ) ( z - x )}

What is expand ((n+1)(6n^4+9n^3+n^2-n))/(30) ?
The solution to expand ((n+1)(6n^4+9n^3+n^2-n))/(30) is (n^5)/5+(n^4)/2+(n^3)/3-n/(30)