# 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?

• 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 $x=c/(a+b)$ – A Googler May 8 '15 at 14:23
• 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 $i_1+i_2+i_3=n? – Iceman May 8 '15 at 14:56 ## 2 Answers For$|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} • That is nice. Thank you. – Iceman May 8 '15 at 15:28 • So when considering the negative integersn$, we can just convert the binomial coefficient using the standard binomial conversion, change$a^{n-k}$to$a^{-1(n+k)}=\frac1{a^{n+k}}$? – Iceman May 8 '15 at 15:57 Yes, see multinomial theorem. It's pretty famous generalization of binomial one. • Thank you, I know of the multinomial theorem, but from what I read, it is for nonnegative integers$n$. My question is dealing with$negative$and$real$numbers. – Iceman May 8 '15 at 15:06 • Well,$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. – Valentin May 8 '15 at 15:33
• You are right. Thank you for your help! – Iceman May 8 '15 at 15:34