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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?

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    $\begingroup$ Binomial series :$(1+x)^k=1 + kx + (k(k-1)/2 )x^2 + ...$ for real $k$ and $x <1$. In your case , suppose $ a+b>c$ . Then $(a+b+c)^n=(a+b)^n(1 + c/(a+b))^k$ . Now you can just use binomial series with $x=c/(a+b)$ $\endgroup$ – A Googler May 8 '15 at 14:23
  • $\begingroup$ Could I also consider that $\binom{n}{i_1,i_2,i_3}=\binom{n}{i_1,i_2,n-i_1-i_2}$, since $i_1+i_2+i_3=n? $\endgroup$ – Iceman May 8 '15 at 14:56
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For $|b+c| < |a|$,

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

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  • $\begingroup$ That is nice. Thank you. $\endgroup$ – Iceman May 8 '15 at 15:28
  • $\begingroup$ So when considering the negative integers $n$, we can just convert the binomial coefficient using the standard binomial conversion, change $a^{n-k}$ to $a^{-1(n+k)}=\frac1{a^{n+k}}$? $\endgroup$ – Iceman May 8 '15 at 15:57
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Yes, see multinomial theorem. It's pretty famous generalization of binomial one.

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    $\begingroup$ Thank you, I know of the multinomial theorem, but from what I read, it is for nonnegative integers $n$. My question is dealing with $negative$ and $real$ numbers. $\endgroup$ – Iceman May 8 '15 at 15:06
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    $\begingroup$ Well, $x^{-n}$ is just $1/x^n$. If you want to compute $(a_1 + a_2 + ... + a_m)^{-m}$, it equals $1/(a_1 + a_2 + ... + a_m)^{m}$. Real numbers follow from this. $\endgroup$ – Valentin May 8 '15 at 15:33
  • $\begingroup$ You are right. Thank you for your help! $\endgroup$ – Iceman May 8 '15 at 15:34

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