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

## Homework Statement

Show that the binomial coefficients ## \binom {-1}{n}=(-1)^n##

## Homework Equations

##\binom{n}{k}=\frac{n!}{(n-k)!k!} \\
##

## 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.

Orodruin
Staff Emeritus
Homework Helper
Gold Member
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.

## Homework Statement

Show that the binomial coefficients ## \binom {-1}{n}=(-1)^n##

## Homework Equations

##\binom{n}{k}=\frac{n!}{(n-k)!k!} \\
##

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

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.

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.

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.

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.
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!