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

The solution to expand ((k+1)(2k+1)(2k+3))/3 is (4k^3)/3+4k^2+(11k)/3+1

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

expand

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

\frac{4 k ^{3}}{3} + 4 k^{2} + \frac{11 k}{3} + 1

Solution steps

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

Expand

(k + 1) (2 k + 1) (2 k + 3) : 4 k^{3} + 12 k^{2} + 11 k + 3

= \frac{4 k ^{3} + 12 k ^{2} + 11 k + 3}{3}

Apply the fraction rule:

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

\frac{4 k ^{3} + 12 k ^{2} + 11 k + 3}{3} = \frac{4 k ^{3}}{3} + \frac{12 k ^{2}}{3} + \frac{11 k}{3} + \frac{3}{3}

= \frac{4 k ^{3}}{3} + \frac{12 k ^{2}}{3} + \frac{11 k}{3} + \frac{3}{3}

Apply rule

\frac{a}{a} = 1

\frac{3}{3} = 1

= \frac{4 k ^{3}}{3} + \frac{12 k ^{2}}{3} + \frac{11 k}{3} + 1

Divide the numbers:

\frac{12}{3} = 4

= \frac{4 k ^{3}}{3} + 4 k^{2} + \frac{11 k}{3} + 1

2 (e^{4} - e)

\frac{8 x ( x ^{2} - 4 )}{( 4 x ^{4} - 32 x ^{2} + 289 ) ^{\frac{1}{2}}}

3 (2 x - 8 = 6)

\frac{d}{d y} ((x) (1 + sin (\frac{π x}{4})))

\frac{f ( x + h ) - f ( x ^{2} - 1 )}{h}

What is expand ((k+1)(2k+1)(2k+3))/3 ?
The solution to expand ((k+1)(2k+1)(2k+3))/3 is (4k^3)/3+4k^2+(11k)/3+1