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

The solution to expand ((n+1)(n+2)(2n+3))/6 is (n^3)/3+(3n^2)/2+(13n)/6+1

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

expand

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

\frac{n ^{3}}{3} + \frac{3 n ^{2}}{2} + \frac{13 n}{6} + 1

Solution steps

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

Expand

(n + 1) (n + 2) (2 n + 3) : 2 n^{3} + 9 n^{2} + 13 n + 6

= \frac{2 n ^{3} + 9 n ^{2} + 13 n + 6}{6}

Apply the fraction rule:

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

\frac{2 n ^{3} + 9 n ^{2} + 13 n + 6}{6} = \frac{2 n ^{3}}{6} + \frac{9 n ^{2}}{6} + \frac{13 n}{6} + \frac{6}{6}

= \frac{2 n ^{3}}{6} + \frac{9 n ^{2}}{6} + \frac{13 n}{6} + \frac{6}{6}

Apply rule

\frac{a}{a} = 1

\frac{6}{6} = 1

= \frac{2 n ^{3}}{6} + \frac{9 n ^{2}}{6} + \frac{13 n}{6} + 1

Cancel

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

= \frac{n ^{3}}{3} + \frac{9 n ^{2}}{6} + \frac{13 n}{6} + 1

Cancel

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

= \frac{n ^{3}}{3} + \frac{3 n ^{2}}{2} + \frac{13 n}{6} + 1

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

(4 + 2 i)^{4}

\frac{( - 2 s + 4 ) ( - s ^{2} + 4 s + 8 )}{( s ^{2} - 4 s + 8 ) ^{3}}

A (- 1, 2) y B (- 3, - 6)

What is expand ((n+1)(n+2)(2n+3))/6 ?
The solution to expand ((n+1)(n+2)(2n+3))/6 is (n^3)/3+(3n^2)/2+(13n)/6+1