Closed form for a sum of binomial coefficients

Question

Let $m,n,r\in\mathbb{N}\cup\{0\}.$ I am interested in finding a closed form for the sum

$$\sum_{i=0}^m{{n+i}\choose{r+i}}.$$

Let $f(m,n,r)$ denote the above sum.

We may make a few trivial observations. For example, $f(0,n,r)={{n}\choose{r}}.$ We also have stuff like $f(m,n,r)=0$ if $n<r$ and $f(m,n,r)=m+1$ if $n=r.$

Is there a closed form for this? By "closed form" I mean I'm looking for an expression for $f(m,n,r)$ in terms of $m,n,$ and $r$ that uses something like binomial coefficients, or the like.

Many times such a closed form doesn't exist, but I think one might just exist here. That's because we have closed forms for stuff like $\sum_{i=0}^n{{n}\choose{i}}$ and $\sum_{i=0}^n{{i}\choose{r}},$ and my sum is just a "combination" of these two.

Maybe there's some slick combinatorial argument for this? Even if there isn't a combinatorial argument that gives a closed form, I'd be interested in a combinatorial interpretation of the above sum.

Answer 1

It's always a good idea to try writing $\binom{a+i}{b+i}$ as $\binom{a+i}{b-a}$ in a sum over $i$; that way, only one parameter of the binomial coefficient varies with $i$. By doing that here, and then re-indexing the sum, we can put it into one of the forms mentioned in the question: $$\sum_{i=0}^m \binom{n+i}{r+i} = \sum_{i=0}^m \binom{n+i}{n-r} = \sum_{j=n}^{n+m} \binom j{n-r}.$$ This is not the same as the sum as $j$ goes from $0$ to $n+m$, which would be $\binom{n+m+1}{n-r+1}$. However, we can write an arbitrary sum like this as a difference of two partial sums starting at $0$: $$ \sum_{j=n}^{n+m} \binom{j}{n-r} = \sum_{j=0}^{n+m} \binom{j}{n-r} - \sum_{j=0}^{n-1} \binom j{n-r} = \binom{n+m+1}{n-r+1} - \binom{n}{n-r+1}. $$