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

The solution to expand (2k+3)/(k(k+1)(k+2)(k+3)) is (2k+3)/(k^4+6k^3+11k^2+6k)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

expand

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

\frac{2 k + 3}{k ^{4} + 6 k ^{3} + 11 k ^{2} + 6 k}

Solution steps

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

Expand

k (k + 1) (k + 2) (k + 3) : k^{4} + 6 k^{3} + 11 k^{2} + 6 k

= \frac{2 k + 3}{k ^{4} + 6 k ^{3} + 11 k ^{2} + 6 k}

\frac{1}{2} cos (2 (x - π))

\frac{14}{( s + 2 ) ( s + 14 )}

((x - 1) \cdot 2^{x + 1}) \cdot 2.5

(- 1, - 3) y (2, - 2)

so l v e (3 x^{2} + 6 x = 0)

What is expand (2k+3)/(k(k+1)(k+2)(k+3)) ?
The solution to expand (2k+3)/(k(k+1)(k+2)(k+3)) is (2k+3)/(k^4+6k^3+11k^2+6k)