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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

expand

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

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

Solution steps

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

Remove parentheses:

(a) = a

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

Expand

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

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

Apply the fraction rule:

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

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

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

Divide the numbers:

\frac{3}{3} = 1

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

c u r l (3 x z, - 7 y, 2 x z^{2})

\frac{x ( 8 - z ) ^{2}}{2}

\frac{d}{d y} ((x (y)) (y^{2} - 1))

(3 i, - 1 j, - 1 k) \cdot (2 i, j, 2 k)

[(x^{2} + 2) (x - 1)]^{^{'}}

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