http://en.wikipedia.org/wiki/Binomial_coefficient#Newton.27s_binomial_series ,

I am trying to prove that

$$ \sum_{\kappa=0}^\infty \binom{\eta + \kappa}{\kappa}x^\kappa = (1 - x)^{-(\kappa+1)}. $$

According to the article, this identity can be derived from the binomial series formula, by applying the identity $$ \binom{n}{k} = (-1)^k\binom{k - n - 1}{k}. $$

To that end, I define $n = \eta + \kappa$ and $k = \kappa$, so that applying the identity, I get

$$ \binom{n}{k} = \binom{\eta + \kappa}{\kappa} = (-1)^\kappa \binom{\eta - \eta - \kappa - 1}{\kappa} = (-1)^\kappa \binom{-\kappa - 1}{\kappa} $$

Now, I do sort of "see", that formally, the binomial series formula should apply. But I don't understand why I am justified in using the formal method. In particular, the binomial coefficients are $0$ when the $n$ term is negative. So I don't really understand the identity

$$ \binom{n}{k} = (-1)^k\binom{k - n - 1}{k}. $$

  • 1
    $\begingroup$ this can easily be proven using differentiation. just note that (1-x)^(-1)= 1+x+x^2+x^3+...... Now differentiate it k times you will observe the result! $\endgroup$ – aditya gupta Jul 11 '20 at 21:36

We define $\binom{x}{k}=x(x-1)\cdots(x-(k-1))/k!$ in any number system where division by $k!$ is possible. So in particular $\binom{x}{k}$ with $x$ a negative integer makes sense and cannot be $0$. The reflection formula then is basically what you get when you factor all of the negative signs out of the product in the numerator in the defining formula for $\binom{x}{k}$ when $x$ is negative.


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