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

The solution to expand (((k+1)+1)(2(k+1)+1)(2(k+1)+3))/3 is (4k^3)/3+8k^2+(47k)/3+10

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

expand

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

\frac{4 k ^{3}}{3} + 8 k^{2} + \frac{47 k}{3} + 10

Solution steps

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

Remove parentheses:

(a) = a

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

Add the numbers:

1 + 1 = 2

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

Expand

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

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

Expand

2 (k + 1) + 3 : 2 k + 5

= \frac{( k + 2 ) ( 2 k + 3 ) ( 2 k + 5 )}{3}

Expand

(k + 2) (2 k + 3) (2 k + 5) : 4 k^{3} + 24 k^{2} + 47 k + 30

= \frac{4 k ^{3} + 24 k ^{2} + 47 k + 30}{3}

Apply the fraction rule:

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

\frac{4 k ^{3} + 24 k ^{2} + 47 k + 30}{3} = \frac{4 k ^{3}}{3} + \frac{24 k ^{2}}{3} + \frac{47 k}{3} + \frac{30}{3}

= \frac{4 k ^{3}}{3} + \frac{24 k ^{2}}{3} + \frac{47 k}{3} + \frac{30}{3}

Divide the numbers:

\frac{24}{3} = 8

= \frac{4 k ^{3}}{3} + 8 k^{2} + \frac{47 k}{3} + \frac{30}{3}

Divide the numbers:

\frac{30}{3} = 10

= \frac{4 k ^{3}}{3} + 8 k^{2} + \frac{47 k}{3} + 10

π (2 - 2 q) (1 - π) (2 + p - q)

cosh (0)

241 (5, - 2) (8, x)

(y^{2} + t y^{2}) y^{'} (t) + t^{2 - y t^{2}}

\frac{d ^{2}}{d t ^{2}} ((v (t)) (2 t^{2} + 4 t + 12))

What is expand (((k+1)+1)(2(k+1)+1)(2(k+1)+3))/3 ?
The solution to expand (((k+1)+1)(2(k+1)+1)(2(k+1)+3))/3 is (4k^3)/3+8k^2+(47k)/3+10