What is expand ((n)(n+1)(2n+7))/6 ?

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

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expand ((n)(n+1)(2n+7))/6

expand

\frac{( n ) ( n + 1 ) ( 2 n + 7 )}{6}

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

Solution steps

\frac{( n ) ( n + 1 ) ( 2 n + 7 )}{6}

Remove parentheses:

(a) = a

= \frac{n ( n + 1 ) ( 2 n + 7 )}{6}

Expand

n (n + 1) (2 n + 7) : 2 n^{3} + 9 n^{2} + 7 n

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

Apply the fraction rule:

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

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

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

Cancel

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

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

Cancel

\frac{9 n ^{2}}{6} : \frac{3 n ^{2}}{2}

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

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

\frac{8 ( x + 1 )}{11}

(x + y - 3 z)^{5}

\frac{1}{( u + 1 ) ( - u + 1 )}

n o t ab l es (x - 5)^{3}

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