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

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

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

expand

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

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

Solution steps

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

Expand

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

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

Apply the fraction rule:

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

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

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

Apply rule

\frac{a}{a} = 1

\frac{6}{6} = 1

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

Cancel

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

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

Cancel

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

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

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

(3 - 2 k)^{2}

\frac{\partial}{\partial x} ((z) (x^{2} + y^{2} + z^{2}))

(2 + 3 i)^{4}

(- 3 + i)^{5}

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