What is expand ((k+1)^4)/4 ?

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

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expand ((k+1)^4)/4

expand

\frac{( k + 1 ) ^{4}}{4}

\frac{k ^{4}}{4} + k^{3} + \frac{3 k ^{2}}{2} + k + \frac{1}{4}

Solution steps

\frac{( k + 1 ) ^{4}}{4}

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

= \frac{k ^{4} + 4 k ^{3} + 6 k ^{2} + 4 k + 1}{4}

Apply the fraction rule:

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

\frac{k ^{4} + 4 k ^{3} + 6 k ^{2} + 4 k + 1}{4} = \frac{k ^{4}}{4} + \frac{4 k ^{3}}{4} + \frac{6 k ^{2}}{4} + \frac{4 k}{4} + \frac{1}{4}

= \frac{k ^{4}}{4} + \frac{4 k ^{3}}{4} + \frac{6 k ^{2}}{4} + \frac{4 k}{4} + \frac{1}{4}

Divide the numbers:

\frac{4}{4} = 1

= \frac{k ^{4}}{4} + k^{3} + \frac{6 k ^{2}}{4} + \frac{4 k}{4} + \frac{1}{4}

Cancel

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

= \frac{k ^{4}}{4} + k^{3} + \frac{3 k ^{2}}{2} + \frac{4 k}{4} + \frac{1}{4}

Divide the numbers:

\frac{4}{4} = 1

= \frac{k ^{4}}{4} + k^{3} + \frac{3 k ^{2}}{2} + k + \frac{1}{4}

(5 x + \frac{4}{x}) (2 x + 3)

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

(1 + x y) d x - (1 + x y^{2}) d y

(3 x - \frac{3}{x})^{5} (2 x^{3} - 1)

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