Is it valid to define $\binom{n}{n+k} = 0$

Question

Is it valid to define

$$\binom{n}{n+k} = 0$$

where $k$ is an integer in $\{k < -n\}\cup\{k > n\}$ ? I couldn't find anything on this notation via a quick google search, but I ran into it in the induction step of the proof that

$$\sum_{k=0}^n{\binom{n}{k}} = 2^n \tag{$\star$}$$

where the following is obtained by Pascal's identity:

\begin{align} \sum_{k=0}^{n+1}{\binom{n+1}{k}} &= \sum_{k=0}^{n+1}{\binom{n}{k-1}} + \sum_{k=0}^{n+1}{\binom{n}{k}} \\ \\ &= \binom{n}{-1} + \sum_{k=1}^{n+1}{\binom{n}{k-1}} + \sum_{k=0}^{n}{\binom{n}{k}} + \binom{n}{n+1} \\ \\ &= 0 + \sum_{k=0}^{n}{\binom{n}{k}} + \sum_{k=0}^{n}{\binom{n}{k}} + 0 \\ \\ &= 2\cdot 2^n \end{align}

where this approach only makes sense if those aforementioned forms are zero. To me it seems to intuitively makes sense, since there are zero ways to do either of those things, since they are impossible.

I'm skeptical that this definition is valid because I saw a proof of $(\star)$where those expressions were avoided by other reasoning.

Answer 1

You may extend the definition of $n \choose r$, by making it 0 if r<0 or r>n.

If you want an intuitive explanation then, look at $(1+x)^n$ which has no $x^{n+1}$ or $x^{-1}$ terms.

Answer 2

It is convenient to extend the definition of binomial coefficients beyond its combinatorial meaning. We can consider $\binom{r}{k}$ and allow $r$ to be a real (or even a complex) number and $k$ to be an integer.

The extended definition is \begin{align*} \binom{r}{k}= \begin{cases} \frac{r(r-1)\cdots (r-k+1)}{k!}&\qquad \text{integer }k\geq 0\\ 0&\qquad \text{integer }k <0 \end{cases}\tag{1} \end{align*}

This definition can be found for instance in section 5.1 Binomial Coefficients, formula (5.1) in Concrete Mathematics by R.L. Graham, D.E. Knuth and O. Patashnik.

From (1) it follows for non-negative integer $n$ and integer $k<-n$ or $k>0$ that $\binom{n}{n+k}=0$.