What is expand (k(k+1)(k+2)(k+3))/4 ?

The solution to expand (k(k+1)(k+2)(k+3))/4 is (k^4)/4+(3k^3)/2+(11k^2)/4+(3k)/2

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expand (k(k+1)(k+2)(k+3))/4

expand

\frac{k ( k + 1 ) ( k + 2 ) ( k + 3 )}{4}

\frac{k ^{4}}{4} + \frac{3 k ^{3}}{2} + \frac{11 k ^{2}}{4} + \frac{3 k}{2}

Solution steps

\frac{k ( k + 1 ) ( k + 2 ) ( k + 3 )}{4}

Expand

k (k + 1) (k + 2) (k + 3) : k^{4} + 6 k^{3} + 11 k^{2} + 6 k

= \frac{k ^{4} + 6 k ^{3} + 11 k ^{2} + 6 k}{4}

Apply the fraction rule:

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

\frac{k ^{4} + 6 k ^{3} + 11 k ^{2} + 6 k}{4} = \frac{k ^{4}}{4} + \frac{6 k ^{3}}{4} + \frac{11 k ^{2}}{4} + \frac{6 k}{4}

= \frac{k ^{4}}{4} + \frac{6 k ^{3}}{4} + \frac{11 k ^{2}}{4} + \frac{6 k}{4}

Cancel

\frac{6 k ^{3}}{4} : \frac{3 k ^{3}}{2}

= \frac{k ^{4}}{4} + \frac{3 k ^{3}}{2} + \frac{11 k ^{2}}{4} + \frac{6 k}{4}

Cancel

\frac{6 k}{4} : \frac{3 k}{2}

= \frac{k ^{4}}{4} + \frac{3 k ^{3}}{2} + \frac{11 k ^{2}}{4} + \frac{3 k}{2}

(1 + tan (x)) tan (2 x)

(e + 3) (e - 5)

(5 - x) (x - \frac{x}{2})

(5 a + b) (5 a - b)

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

What is expand (k(k+1)(k+2)(k+3))/4 ?
The solution to expand (k(k+1)(k+2)(k+3))/4 is (k^4)/4+(3k^3)/2+(11k^2)/4+(3k)/2