What is expand ((k^2+k)/2)^2 ?

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

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expand ((k^2+k)/2)^2

expand

(\frac{k ^{2} + k}{2})^{2}

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

Solution steps

(\frac{k ^{2} + k}{2})^{2}

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{( k ^{2} + k ) ^{2}}{2 ^{2}}

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

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

2^{2} = 4

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

Apply the fraction rule:

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

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

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

Cancel

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

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

(π + i)^{2}

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

(x^{2} - y) y^{''} (x)

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

\frac{( z + J ^{2} ) ( z - 14 ) 5}{z - 14 + J ^{2}}

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