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

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

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

expand

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

\frac{k ^{3}}{3} + k^{2} + \frac{2 k}{3} + 2 k + 2

Solution steps

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

Remove parentheses:

(a) = a

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

Expand

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

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

Apply the fraction rule:

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

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

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

Divide the numbers:

\frac{3}{3} = 1

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

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- \frac{3 ( 4 x ^{3} + 6 x ^{2} h + 4 x h ^{2} + h ^{3} )}{x ^{4} ( x + h )}

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(n + 1) e^{\frac{1}{n}}

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