General expanded form of $(x+y+z)^k$

Question

(hope it doesn't seem so weird), I'm looking for a general expanded form of $(x+y+z)^k, k\in\mathbb{N}$.

$k=1: x+y+z$

$k=2: x^2+y^2+z^2+2xy+2xz+2yz$

$k=3: x^3+y^3+z^3+3xy^2+3xz^2+3yz^2+3x^2y+3x^2z+3y^2z+6xyz$

$k=4: x^4+y^4+z^4+4xy^3+4x^3y+4xz^3+4x^3z+4yz^3 +4y^3z+6x^2y^2+6y^2z^2+6x^2z^2+12x^2yz+12xy^2z+12xyz^2$

The elements are obviously determined by combinations of their powers, whose sum is always $k$. I just cannot find the algorithm for element's constants

Related: How to expand $(a_0+a_1x+a_2x^2+...a_nx^n)^2$? – Dave L. Renfro Jun 17 '14 at 14:42 — Dave L. Renfro, Jun 17 '14 at 14:42

beep-boop · Answer 1 · 2014-06-17T14:48:54.017

6

Hint: $(x+y+z)^k \equiv[\color{red}x+\color{green}{(y+z)}]^k$.

Now use the binomial formula for $(\color{red}a+\color{green}b)^k$.

edited Jun 17 '14 at 14:48

answered Jun 17 '14 at 14:43

beep-boop

11,595

For the specific question, I much prefer this answer to mine. – JP McCarthy Jun 17 '14 at 14:45
1

@JpMcCarthy http://imgur.com/iOFLp – beep-boop Jun 17 '14 at 14:46

score 4 · Answer 2 · answered Jun 17 '14 at 14:36

4

It is the Multinomial Theorem.

answered Jun 17 '14 at 14:36

JP McCarthy

8,420

score 4 · Accepted Answer · answered Jun 17 '14 at 14:52

4

Note that just as you can use Pascal's Triangle for binomials, you can use Pascal's Pyramid for trinomials. Otherwise, you can use the Multinomial Theorem as Jp McCarthy suggested

Pascals' Pyramid

answered Jun 17 '14 at 14:52

scrblnrd3

488

General expanded form of $(x+y+z)^k$

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