What is expand ((n+1)(n+2)(2n+9))/6 ?

The solution to expand ((n+1)(n+2)(2n+9))/6 is (n^3)/3+(5n^2)/2+(31n)/6+3

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expand ((n+1)(n+2)(2n+9))/6

expand

\frac{( n + 1 ) ( n + 2 ) ( 2 n + 9 )}{6}

\frac{n ^{3}}{3} + \frac{5 n ^{2}}{2} + \frac{31 n}{6} + 3

Solution steps

\frac{( n + 1 ) ( n + 2 ) ( 2 n + 9 )}{6}

Expand

(n + 1) (n + 2) (2 n + 9) : 2 n^{3} + 15 n^{2} + 31 n + 18

= \frac{2 n ^{3} + 15 n ^{2} + 31 n + 18}{6}

Apply the fraction rule:

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

\frac{2 n ^{3} + 15 n ^{2} + 31 n + 18}{6} = \frac{2 n ^{3}}{6} + \frac{15 n ^{2}}{6} + \frac{31 n}{6} + \frac{18}{6}

= \frac{2 n ^{3}}{6} + \frac{15 n ^{2}}{6} + \frac{31 n}{6} + \frac{18}{6}

Cancel

\frac{2 n ^{3}}{6} : \frac{n ^{3}}{3}

= \frac{n ^{3}}{3} + \frac{15 n ^{2}}{6} + \frac{31 n}{6} + \frac{18}{6}

Cancel

\frac{15 n ^{2}}{6} : \frac{5 n ^{2}}{2}

= \frac{n ^{3}}{3} + \frac{5 n ^{2}}{2} + \frac{31 n}{6} + \frac{18}{6}

Divide the numbers:

\frac{18}{6} = 3

= \frac{n ^{3}}{3} + \frac{5 n ^{2}}{2} + \frac{31 n}{6} + 3

\frac{2}{x ( x ^{2} - 16 )}

f (x) = \frac{2 x ^{5} + 28 x ^{3} - 6 x}{( x ^{2} - 1 ) ^{4}}

\frac{100 a ^{2} - 20 a + 1}{100}

(y^{'} (x) + (1 - sec (x)) y^{2}) tan (x)

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

What is expand ((n+1)(n+2)(2n+9))/6 ?
The solution to expand ((n+1)(n+2)(2n+9))/6 is (n^3)/3+(5n^2)/2+(31n)/6+3