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

The solution to expand ((k+1)^2(k+2)^2)/4+(k+1) is 4k+(k^4)/4+(3k^3)/2+(13k^2)/4+2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

expand

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

4 k + \frac{k ^{4}}{4} + \frac{3 k ^{3}}{2} + \frac{13 k ^{2}}{4} + 2

Solution steps

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

Remove parentheses:

(a) = a

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

\frac{( k + 1 ) ^{2} ( k + 2 ) ^{2}}{4} = \frac{k ^{4} + 6 k ^{3} + 13 k ^{2} + 12 k + 4}{4}

= \frac{k ^{4} + 6 k ^{3} + 13 k ^{2} + 12 k + 4}{4} + k + 1

Apply the fraction rule:

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

\frac{k ^{4} + 6 k ^{3} + 13 k ^{2} + 12 k + 4}{4} = \frac{k ^{4}}{4} + \frac{6 k ^{3}}{4} + \frac{13 k ^{2}}{4} + \frac{12 k}{4} + \frac{4}{4}

= \frac{k ^{4}}{4} + \frac{6 k ^{3}}{4} + \frac{13 k ^{2}}{4} + \frac{12 k}{4} + \frac{4}{4} + k + 1

Apply rule

\frac{a}{a} = 1

\frac{4}{4} = 1

= \frac{k ^{4}}{4} + \frac{6 k ^{3}}{4} + \frac{13 k ^{2}}{4} + \frac{12 k}{4} + 1 + k + 1

Cancel

\frac{6 k ^{3}}{4} : \frac{3 k ^{3}}{2}

= \frac{k ^{4}}{4} + \frac{3 k ^{3}}{2} + \frac{13 k ^{2}}{4} + \frac{12 k}{4} + 1 + k + 1

Divide the numbers:

\frac{12}{4} = 3

= \frac{k ^{4}}{4} + \frac{3 k ^{3}}{2} + \frac{13 k ^{2}}{4} + 3 k + 1 + k + 1

Group like terms

= 3 k + k + \frac{k ^{4}}{4} + \frac{3 k ^{3}}{2} + \frac{13 k ^{2}}{4} + 1 + 1

Add similar elements:

3 k + k = 4 k

= 4 k + \frac{k ^{4}}{4} + \frac{3 k ^{3}}{2} + \frac{13 k ^{2}}{4} + 1 + 1

Add the numbers:

1 + 1 = 2

= 4 k + \frac{k ^{4}}{4} + \frac{3 k ^{3}}{2} + \frac{13 k ^{2}}{4} + 2

- \frac{4}{s ^{2} ( s - 6 )}

\frac{5 x + 3}{( 2 x + 1 ) ( x + 1 ) ^{2}}

(5 x - \frac{1}{2 x ^{3}})^{8}

(- 5 + 5 i)^{5}

3 \cdot (4 - \frac{x}{9})

What is expand ((k+1)^2(k+2)^2)/4+(k+1) ?
The solution to expand ((k+1)^2(k+2)^2)/4+(k+1) is 4k+(k^4)/4+(3k^3)/2+(13k^2)/4+2