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

The solution to expand (k(2k+1)(4k+1))/3+(2(k+1))^2 is 6k^2+8k+(8k^3)/3+k/3+4

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expand (k(2k+1)(4k+1))/3+(2(k+1))^2

expand

\frac{k ( 2 k + 1 ) ( 4 k + 1 )}{3} + (2 (k + 1))^{2}

6 k^{2} + 8 k + \frac{8 k ^{3}}{3} + \frac{k}{3} + 4

Solution steps

\frac{k ( 2 k + 1 ) ( 4 k + 1 )}{3} + (2 (k + 1))^{2}

\frac{k ( 2 k + 1 ) ( 4 k + 1 )}{3} = \frac{8 k ^{3} + 6 k ^{2} + k}{3}

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

= \frac{8 k ^{3} + 6 k ^{2} + k}{3} + 2^{2} (k^{2} + 2 k + 1)

2^{2} = 4

= \frac{8 k ^{3} + 6 k ^{2} + k}{3} + 4 (k^{2} + 2 k + 1)

Expand

4 (k^{2} + 2 k + 1) : 4 k^{2} + 8 k + 4

= \frac{8 k ^{3} + 6 k ^{2} + k}{3} + 4 k^{2} + 8 k + 4

Apply the fraction rule:

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

\frac{8 k ^{3} + 6 k ^{2} + k}{3} = \frac{8 k ^{3}}{3} + \frac{6 k ^{2}}{3} + \frac{k}{3}

= \frac{8 k ^{3}}{3} + \frac{6 k ^{2}}{3} + \frac{k}{3} + 4 k^{2} + 8 k + 4

Divide the numbers:

\frac{6}{3} = 2

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

Group like terms

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

Add similar elements:

2 k^{2} + 4 k^{2} = 6 k^{2}

= 6 k^{2} + 8 k + \frac{8 k ^{3}}{3} + \frac{k}{3} + 4

(x - \frac{5}{4})^{2} + 3

(x sin (θ) + y cos (θ))^{2}

(\frac{( - 1 - 41 )}{2})^{4}

\frac{4 x ( x ^{3} + 2 ) + 3 x ^{2}}{2 ( x ^{3} + 2 ) ^{\frac{1}{2}}}

(62500) π (\frac{2}{3} - \frac{3 3}{8})

What is expand (k(2k+1)(4k+1))/3+(2(k+1))^2 ?
The solution to expand (k(2k+1)(4k+1))/3+(2(k+1))^2 is 6k^2+8k+(8k^3)/3+k/3+4