1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Prove the binomial coefficients are (-1)^n

  1. May 23, 2015 #1
    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} \\
    =\frac{-1}{(-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. jcsd
  3. May 23, 2015 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member
    2017 Award

    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.
  4. May 23, 2015 #3
    That is not the definition of the binomial coefficient for negative ##n##.
  5. May 23, 2015 #4
    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.
  6. May 23, 2015 #5
    Yes there is. Just use the general definition of the binomial:
    [tex]\binom{n}{k} = \frac{n(n-1)(n-2)...(n-k+1)}{k!}[/tex]
    Plug it in and calculate a little bit.
  7. May 23, 2015 #6
    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!} \\
    Thanks that formula helped a lot!
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted