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

The solution to expand ((n+1)(n+3)(2n+9))/6 is (n^3)/3+(17n^2)/6+7n+9/2

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expand ((n+1)(n+3)(2n+9))/6

expand

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

\frac{n ^{3}}{3} + \frac{17 n ^{2}}{6} + 7 n + \frac{9}{2}

Solution steps

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

Expand

(n + 1) (n + 3) (2 n + 9) : 2 n^{3} + 17 n^{2} + 42 n + 27

= \frac{2 n ^{3} + 17 n ^{2} + 42 n + 27}{6}

Apply the fraction rule:

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

\frac{2 n ^{3} + 17 n ^{2} + 42 n + 27}{6} = \frac{2 n ^{3}}{6} + \frac{17 n ^{2}}{6} + \frac{42 n}{6} + \frac{27}{6}

= \frac{2 n ^{3}}{6} + \frac{17 n ^{2}}{6} + \frac{42 n}{6} + \frac{27}{6}

Cancel

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

= \frac{n ^{3}}{3} + \frac{17 n ^{2}}{6} + \frac{42 n}{6} + \frac{27}{6}

Divide the numbers:

\frac{42}{6} = 7

= \frac{n ^{3}}{3} + \frac{17 n ^{2}}{6} + 7 n + \frac{27}{6}

Cancel

\frac{27}{6} : \frac{9}{2}

= \frac{n ^{3}}{3} + \frac{17 n ^{2}}{6} + 7 n + \frac{9}{2}

\frac{4 x ^{2}}{( 2 x ^{3} - 11 ) ^{\frac{1}{3}} \cdot ( 2 x - 1 )}

\frac{1}{z ( 1 - z ) ^{3}}

(e^{5 x} + 4 e^{- 3 x})^{2}

\frac{( 13 - 11 ) 13 + 11}{79}

(- 2 + 4 i) (- 2 + 4 i)^{'}

What is expand ((n+1)(n+3)(2n+9))/6 ?
The solution to expand ((n+1)(n+3)(2n+9))/6 is (n^3)/3+(17n^2)/6+7n+9/2