combinatorics - Trinomial Theorem for negative exponents

I just learned of binomial theorem for negative integers (or in that case any real $n$). Does such a theorem exist for the trinomial theorem $$(a+b+c)^n$$ and has there been work done?

I would think that it could logically be extended in the same way as the binomial. You could look at $$(a+b+c)^{-n}$$ The first step would be defining the trinomial coefficient. So $$\binom{-n}{i_1,i_2,i_3}=\frac{(-n)(-n-1)(-n-2)...}{i_1!i_2!i_3!}$$

But this really doesn't make sense to me. It seems to work in the binomial case since you have $(n-k)!$ in the denominator. For example, for $n=6, i_1=1, i_2=2, i_3=3$, then $$\binom{-6}{1,2,3}=\frac{-6(-6-1)(-6-2)(-6-3)}{1!2!3!}=(-1)^4\frac{6\cdot 7\cdot 8\cdot 9}{1!2!3!}=(-1)\frac{9!}{1!2!3!5!}$$

But this would not be the only interpretation, because which $i_j$ would you expand to? Any insight?

EDIT: Reconsidering: If I have

$$\binom{n}{i_1,i_2,i_3}$$ Since it is true that $n=i_1+i_2+i_3$, I can write it as $$\binom{n}{i_1,i_2,i_3}=\binom{n}{i_1,i_2,n-i_1-i_2}$$

$$\binom{n}{i_1,i_2,n-i_1-i_2}=\frac{n!}{i_1!i_2!(n-i_1-i_2)!}=\frac{n(n-1)n-2)...(n-i_1-i_2+1)}{i_1!i_2!}$$ Now considering $negative$ $n$, $$\binom{-n}{i_1,i_2,n-i_1-i_2}=\frac{-n(-n-1)(n-2)...(-n-i_1-i_2+1)}{i_1!i_2!}$$ $$=(-1)^{i_1+i_2}\frac{n(n+1)(n-2)...(n+i_1+i_2-1)}{i_1!i_2!}$$ $$=(-1)^{i_1+i_2}\frac{(n+i_1+i_2-1)!}{i_1!i_2!(n-1)!}$$ $$=(-1)^{i_2+i_3}\binom{n+i_1+i_2-1}{i_1,i_2,n-1}$$

Does this make sense?

Answer

For $|b+c| < |a|$,

$$\eqalign{(a+b+c)^{n} &= \sum_{k=0}^\infty {n \choose k} a^{n-k} (b+c)^k\cr & = \sum_{k=0}^\infty \sum_{j=0}^k {n \choose k} {k \choose j} a^{n-k} b^{k-j} c^j\cr}$$