Is there a formula for $(x+y+z)^n$?

Question

We have that $$(x+y)^n=\sum_{k=0}^n\binom{n}{k}x^ky^{n-k},$$ but is there such a formula for $(x+y+z)^n$ ?

Answer 1

$$(x+y+z)^n=([x+y]+z)^n=\sum_{k=0}^n\binom{n}{k}(x+y)^kz^{n-k}=\sum_{k=0}^n\binom nk\left[\sum_{j=0}^k\binom{k}{j}x^jy^{k-j}\right]z^{n-k}$$

Answer 2

What you are looking for is the multinomial theorem.

In your case (3 variables), a bit of combinatronics yields: $$ (x+y+z)^n=\sum_{k+l+m=n} \frac{n!}{k!l!m!}x^ky^lz^m $$

Answer 3

There is::

$$(x+y+z)^n= \!\!\sum_{\substack{(i,j,k)\in\mathbf N^3\\i+j+k=n}}\!\frac{n!}{i!\,j!\,k!} x^i y^j z^k.$$ It can be extended to sums of more than $3$ terms (multinomial formula): $$(x_1+x_2+\dots+x_r)^n= \!\!\sum_{\substack{(i_1,i_2,\dots, i_r)\in\mathbf N^r\\i_1+i_2+\dots+i_r=n}}\!\frac{n!}{i_1!\,i_2!\,\dotsm i_r!}\, x_1^{i_1} x_2^{i_2}\dotsm \,x_r^{i_r}.$$