# Prove the binomial coefficients are (-1)^n

1. May 23, 2015

### Potatochip911

1. The problem statement, all variables and given/known data
Show that the binomial coefficients $\binom {-1}{n}=(-1)^n$

2. Relevant equations
$\binom{n}{k}=\frac{n!}{(n-k)!k!} \\$
3. The attempt at a solution
$-1!=(-1)\cdot 1! \\ -1!=-1 \\ -2!=(-1)^2 \cdot 2! \\ -n!=(-1)^n \cdot n!\\ \mbox{for n=0} \\ LHS=\binom{-1}{0}=\frac{-1!}{(-1!)(0!)}=1 \\ RHS=(-1)^0=1 \\ LHS=RHS \space \mbox{When n=0 therefore it is valid for n=0} \\ \mbox{Assuming it holds for some integer k}\\ \binom{-1}{k}=(-1)^k \\ \mbox{Now trying to prove for k+1} \\ \binom{-1}{k+1}=(-1)^{k+1} \\ LHS=\binom{-1}{k+1} =\frac{-1!}{[(-1-(k+1)]!(k+1)!}\\ =\frac{-1}{(-k-2)!(k+1)!} \\ =\frac{-1}{(-1)^{k+2}(k+2)!(k+1)!}\\ =\frac{1}{(-1)^{k+1}(k+2)!(k+1)!}$
So I obviously did something wrong since I didn't get both sides being equal, it probably has to do with how I treated the negative factorials.

2. May 23, 2015

### Orodruin

Staff Emeritus
Factorials are not defined for negative numbers. If you want an analytic continuation you get the Gamma function, but it has poles at negative integers. It is possible you could define it as a limit when $n \to -1$. In that case, you would get a ratio of residues.

3. May 23, 2015

### micromass

That is not the definition of the binomial coefficient for negative $n$.

4. May 23, 2015

### Potatochip911

Is there a relatively simple method to proving this? I've only taken calc 1, calc 2, and linear algebra so I don't have very much knowledge.

5. May 23, 2015

### micromass

Yes there is. Just use the general definition of the binomial:
$$\binom{n}{k} = \frac{n(n-1)(n-2)...(n-k+1)}{k!}$$
Plug it in and calculate a little bit.

6. May 23, 2015

### Potatochip911

Okay so I think I got it now, $$\binom{n}{k}=\frac{n(n-1)(n-2)\cdots(n-k+2)(n-k+1)}{k!} \\ \binom{-1}{n}=\frac{(-1)(-2)(-3)\cdots (-n+1)(-n)}{n!} \\ \binom{-1}{n}=\frac{(-1)^n(1)(2)(3)\cdots (n-1)(n)}{n!} \\ \binom{-1}{n}=(-1)^n$$
Thanks that formula helped a lot!