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

The solution to expand ((k+1)(k+2)(2k+9))/6 is (k^3)/3+(5k^2)/2+(31k)/6+3

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expand ((k+1)(k+2)(2k+9))/6

expand

\frac{( k + 1 ) ( k + 2 ) ( 2 k + 9 )}{6}

\frac{k ^{3}}{3} + \frac{5 k ^{2}}{2} + \frac{31 k}{6} + 3

Solution steps

\frac{( k + 1 ) ( k + 2 ) ( 2 k + 9 )}{6}

Expand

(k + 1) (k + 2) (2 k + 9) : 2 k^{3} + 15 k^{2} + 31 k + 18

= \frac{2 k ^{3} + 15 k ^{2} + 31 k + 18}{6}

Apply the fraction rule:

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

\frac{2 k ^{3} + 15 k ^{2} + 31 k + 18}{6} = \frac{2 k ^{3}}{6} + \frac{15 k ^{2}}{6} + \frac{31 k}{6} + \frac{18}{6}

= \frac{2 k ^{3}}{6} + \frac{15 k ^{2}}{6} + \frac{31 k}{6} + \frac{18}{6}

Cancel

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

= \frac{k ^{3}}{3} + \frac{15 k ^{2}}{6} + \frac{31 k}{6} + \frac{18}{6}

Cancel

\frac{15 k ^{2}}{6} : \frac{5 k ^{2}}{2}

= \frac{k ^{3}}{3} + \frac{5 k ^{2}}{2} + \frac{31 k}{6} + \frac{18}{6}

Divide the numbers:

\frac{18}{6} = 3

= \frac{k ^{3}}{3} + \frac{5 k ^{2}}{2} + \frac{31 k}{6} + 3

lo g_{4} (3 x^{2} y^{4})

- ln 1 + cos^{2} (7.5)

\frac{8 x}{( x + 2 ) ( x - 2 )}

(λ - e^{- w} μ) (e^{w} - 1)

\frac{x - 4}{( x + 1 ) ( x + 2 )}

What is expand ((k+1)(k+2)(2k+9))/6 ?
The solution to expand ((k+1)(k+2)(2k+9))/6 is (k^3)/3+(5k^2)/2+(31k)/6+3