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

The solution to expand ((k+1)(k+2)(k+3))/6 is (k^3)/6+k^2+(11k)/6+1

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

expand

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

\frac{k ^{3}}{6} + k^{2} + \frac{11 k}{6} + 1

Solution steps

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

Expand

(k + 1) (k + 2) (k + 3) : k^{3} + 6 k^{2} + 11 k + 6

= \frac{k ^{3} + 6 k ^{2} + 11 k + 6}{6}

Apply the fraction rule:

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

\frac{k ^{3} + 6 k ^{2} + 11 k + 6}{6} = \frac{k ^{3}}{6} + \frac{6 k ^{2}}{6} + \frac{11 k}{6} + \frac{6}{6}

= \frac{k ^{3}}{6} + \frac{6 k ^{2}}{6} + \frac{11 k}{6} + \frac{6}{6}

Apply rule

\frac{a}{a} = 1

\frac{6}{6} = 1

= \frac{k ^{3}}{6} + \frac{6 k ^{2}}{6} + \frac{11 k}{6} + 1

Divide the numbers:

\frac{6}{6} = 1

= \frac{k ^{3}}{6} + k^{2} + \frac{11 k}{6} + 1

(p + \frac{a}{v ^{2}}) (v - b)

cos ((5 - n) π)

- \frac{4 x y + 8 y - 6}{( x + 2 ) ^{2}}

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

(15 x^{3} + 7 x^{2} - 11 x + 17) d x

What is expand ((k+1)(k+2)(k+3))/6 ?
The solution to expand ((k+1)(k+2)(k+3))/6 is (k^3)/6+k^2+(11k)/6+1