What is expand ([n(n+1)(2(n)+1)])/6-n ?

The solution to expand ([n(n+1)(2(n)+1)])/6-n is (n^3)/3+(n^2)/2+n/6-n

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expand ([n(n+1)(2(n)+1)])/6-n

expand

\frac{[ n ( n + 1 ) ( 2 ( n ) + 1 ) ]}{6} - n

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

Solution steps

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

Remove parentheses:

(a) = a

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

Expand

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

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

Apply the fraction rule:

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

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

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

Cancel

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

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

Cancel

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

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

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