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

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

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

expand

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

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

Solution steps

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

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

= (\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}

(2 sin (x) + 1) (2 sin (x) - 1)

(e^{t} + t e^{t})^{2}

(e^{- x} - 2)^{2}

2 x l e f t (3 x^{2} - 5 x + 1)

4^{2}

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