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

In summary, the conversation was about finding the expanded form of (x+y+z)^k for different values of k. The expanded form for k=1 is x+y+z, for k=2 is x^2+y^2+z^2+2xy+2xz+2yz, for k=3 is x^3+y^3+z^3+3xy^2+3xz^2+3yz^2+3x^2y+3x^2z+3y^2z+6xyz, and for k=4 is x^4+y^4+z^4+4xy^3+4x^3y+4xz^3+4x^3z+
  • #1
sludger13
83
0
Hi,
(hope it doesn't seem so weird),
I'm looking for a general expanded form of
[itex](x+y+z)^{k}[/itex], [itex]k\in N[/itex]

[itex]k=1[/itex]:
[itex]x+y+z[/itex]

[itex]k=2[/itex]:
[itex]x^{2}+y^{2}+z^{2}+2xy+2xz+2yz[/itex]

[itex]k=3[/itex]:
[itex]x^{3}+y^{3}+z^{3}+3xy^{2}+3xz^{2}+3yz^{2}+3x^{2}y+3x^{2}z+3y^{2}z+6xyz[/itex]

[itex]k=4[/itex]:
[itex]x^{4}+y^{4}+z^{4}+4xy^{3}+4x^{3}y+4xz^{3}+4x^{3}z+4yz^{3}[/itex]
[itex]+4y^{3}z+6x^{2}y^{2}+6y^{2}z^{2}+6x^{2}z^{2}+12x^{2}yz+12xy^{2}z+12xyz^{2}[/itex]

The elements are obviously determined by combinations of their powers, which sum is always [itex]k[/itex].
I just cannot find the algorithm for element's constants.
 
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  • #3
Thanks, I completely forgot to check out the factorials :)
 

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

What is the expanded form of (x+y+z)^k?

The expanded form of (x+y+z)^k is given by the binomial theorem, which states that (x+y+z)^k = ∑ (k choose n) * x^(k-n) * y^n * z^(n).

What is the coefficient of x^2 in the expansion of (x+y+z)^k?

The coefficient of x^2 in the expansion of (x+y+z)^k is (k choose 2) * y^(k-2) * z^2.

How many terms are in the expansion of (x+y+z)^k?

There are (k+1) terms in the expansion of (x+y+z)^k.

What is the degree of the expansion of (x+y+z)^k?

The degree of the expansion of (x+y+z)^k is k, as the highest power of any term is x^k.

Can the binomial theorem be used for any value of k?

Yes, the binomial theorem can be used for any positive integer value of k.

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