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

The solution to expand (((k+1)(k+1+1)(2(k+1)+1)))/6 is (k^3)/3+(3k^2)/2+(13k)/6+1

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

expand

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

\frac{k ^{3}}{3} + \frac{3 k ^{2}}{2} + \frac{13 k}{6} + 1

Solution steps

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

Add the numbers:

1 + 1 = 2

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

Expand

2 (k + 1) + 1 : 2 k + 3

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

Expand

(k + 1) (k + 2) (2 k + 3) : 2 k^{3} + 9 k^{2} + 13 k + 6

= \frac{2 k ^{3} + 9 k ^{2} + 13 k + 6}{6}

Apply the fraction rule:

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

\frac{2 k ^{3} + 9 k ^{2} + 13 k + 6}{6} = \frac{2 k ^{3}}{6} + \frac{9 k ^{2}}{6} + \frac{13 k}{6} + \frac{6}{6}

= \frac{2 k ^{3}}{6} + \frac{9 k ^{2}}{6} + \frac{13 k}{6} + \frac{6}{6}

Apply rule

\frac{a}{a} = 1

\frac{6}{6} = 1

= \frac{2 k ^{3}}{6} + \frac{9 k ^{2}}{6} + \frac{13 k}{6} + 1

Cancel

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

= \frac{k ^{3}}{3} + \frac{9 k ^{2}}{6} + \frac{13 k}{6} + 1

Cancel

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

= \frac{k ^{3}}{3} + \frac{3 k ^{2}}{2} + \frac{13 k}{6} + 1

\frac{25}{( 35 s + 1 ) ( 80 s + 1 )}

(3 x - 2)^{2}

(1 + tan (x)) (sec (x) tan (x))

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

(π + i)^{2}

What is expand (((k+1)(k+1+1)(2(k+1)+1)))/6 ?
The solution to expand (((k+1)(k+1+1)(2(k+1)+1)))/6 is (k^3)/3+(3k^2)/2+(13k)/6+1