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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

expand

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

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

Solution steps

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

Expand

2 k (k + 1) (2 k + 1) : 4 k^{3} + 6 k^{2} + 2 k

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

Apply the fraction rule:

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

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

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

Divide the numbers:

\frac{6}{3} = 2

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

(x - y + 3) d x + (3 x + y + 1) d y

\frac{( ( t - 9 ) ^{2} )}{t ^{5}}

y^{'} (t) + (1 + 3 y)^{3}

\frac{c}{s ( s + k )}

(x^{3} - 2 x) d x

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