Sums of products of binomial coefficients often have simpler expression which do not involve any summation. Examples are the elementary $$\sum_{i=0}^k\binom{a}{i}\binom{b}{k-i}=\binom{a+b}{k}$$ or the more complicated $$\sum_{i=0}^{\text{min(a,b)}}\binom{x+y+i}{i}\binom{y}{a-i}\binom{x}{b-i}=\binom{x+a}{b}\binom{y+b}{a}$$ or many others like Vandermonde-Chu identity etc.
I am looking for a similar formula for the following sum $$S(a,b):=\sum_{i=0}^k\binom{a}{k-i}\binom{b+i}{i}$$ where here $k\le a$ and everybody is a positive integer. This expression has arisen to me as a coefficient of $(1+q)^a(1-q)^{-1-b}$ and I am interested in computing determinants of such numbers. Therefore I am not interested in a "generating function" answer, but rather in an expression that does not involve a summation, like the ones before. Alternatively, a natural combinatorial interpretation can also be useful, in order to use Viennot's theory of binomial determinants.
Anybody here familiar with combinatorics can give me any help? Thank you in advance.
Best Answer
Lacking reputation, I am unable to comment. I will add, however, that a closed ("summation-signless") expression is not forthcoming. I will adjust your notation modestly and write $$ S_k(a,b) = \sum_{0\leq i \leq k}{a\choose k-i}{b+i\choose i} $$ for $a,b,k\in \mathbb{N}$ with $k\leq a$.
Let us also write $S_k(a,0) = \sum_{0\leq i \leq k}{a\choose k-i} = \sum_{0\leq i \leq k}{a\choose i}$; when $k<a$, it is well-known that $S_k(a,0)$ (the proper prefix sum of the $a$-th row of Pascal's Triangle) does not admit a closed form.
We begin by writing \begin{alignat}{2} S_k(a,1) &=\sum_{0\leq i \leq k}{a\choose k-i}(i+1)\notag \\ &= \sum_{0\leq i \leq k}{a\choose k-i}i+ S_k(a,0) \notag\\ &= \sum_{1\leq i \leq k}{a\choose k-i}i+ S_k(a,0) \notag\\ &= \sum_{0\leq i \leq k-1}{a\choose k-(i+1)}(i+1)+ S_k(a,0) \notag\\ &= S_{k-1}(a,1)+ S_k(a,0). \notag\\ \end{alignat}
Thus, $S_k(a,0) = S_k(a,1) - S_{k-1}(a,1)$; a closed form for both $S_{k}(a,1)$ and $S_{k-1}(a,1)$ would lead to a closed difference expression for $S_k(a,0)$.