What is expand (k(k+1)^3(2k+1))/6 ?

The solution to expand (k(k+1)^3(2k+1))/6 is (k^5)/3+(7k^4)/6+(3k^3)/2+(5k^2)/6+k/6

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expand (k(k+1)^3(2k+1))/6

expand

\frac{k ( k + 1 ) ^{3} ( 2 k + 1 )}{6}

\frac{k ^{5}}{3} + \frac{7 k ^{4}}{6} + \frac{3 k ^{3}}{2} + \frac{5 k ^{2}}{6} + \frac{k}{6}

Solution steps

\frac{k ( k + 1 ) ^{3} ( 2 k + 1 )}{6}

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

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

Expand

k (k^{3} + 3 k^{2} + 3 k + 1) (2 k + 1) : 2 k^{5} + 7 k^{4} + 9 k^{3} + 5 k^{2} + k

= \frac{2 k ^{5} + 7 k ^{4} + 9 k ^{3} + 5 k ^{2} + k}{6}

Apply the fraction rule:

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

\frac{2 k ^{5} + 7 k ^{4} + 9 k ^{3} + 5 k ^{2} + k}{6} = \frac{2 k ^{5}}{6} + \frac{7 k ^{4}}{6} + \frac{9 k ^{3}}{6} + \frac{5 k ^{2}}{6} + \frac{k}{6}

= \frac{2 k ^{5}}{6} + \frac{7 k ^{4}}{6} + \frac{9 k ^{3}}{6} + \frac{5 k ^{2}}{6} + \frac{k}{6}

Cancel

\frac{2 k ^{5}}{6} : \frac{k ^{5}}{3}

= \frac{k ^{5}}{3} + \frac{7 k ^{4}}{6} + \frac{9 k ^{3}}{6} + \frac{5 k ^{2}}{6} + \frac{k}{6}

Cancel

\frac{9 k ^{3}}{6} : \frac{3 k ^{3}}{2}

= \frac{k ^{5}}{3} + \frac{7 k ^{4}}{6} + \frac{3 k ^{3}}{2} + \frac{5 k ^{2}}{6} + \frac{k}{6}

\frac{2 x ^{2} - 4 x - 16}{3 ( x + 3 )}

\frac{5 ( 5 + 809 )}{49}

\frac{9 ( 27 + 753 )}{2}

\frac{4 ( y - x )}{5}

\frac{5}{2 z \cdot ( 2 x - 8 y )}

What is expand (k(k+1)^3(2k+1))/6 ?
The solution to expand (k(k+1)^3(2k+1))/6 is (k^5)/3+(7k^4)/6+(3k^3)/2+(5k^2)/6+k/6