Vandermonde-like sum

Question

I know a Vandermonde's identity as

$$ \sum_{i=0}^c {a \choose i} {b \choose c-i} = {a+b \choose c} $$ $$ a, b, c \in \mathbb{N} $$

I am looking for a way to simplify these expressions: $$ \sum_{i=0}^c {a \choose i} {b \choose c-i}(-1)^i\tag{1} $$ $$\sum_{i=x_1}^{x_2} {a \choose i} {b \choose c-i}(-1)^i\quad (x_1\ge 0,\; x_2\le a).\tag{2} $$

You might want to look at generating functions. Again, most combinatorial identities may be solved by the Snake Oil method. — Torsten Hĕrculĕ Cärlemän, Jul 19 '13 at 10:45

score 2 · Answer 1 · answered Dec 27 '13 at 13:41

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Except for trivial cases (if you sum just one term in your second expression, say), your sums do not admit a simple closed formula like the Vandermonde identity. More precisely, you can find a recurrence relation for your expression and use Petkovšek's algorithm to prove that it does not admit a hypergeometric solution.

answered Dec 27 '13 at 13:41

Phira

21,206

2

In particular, let $f(c)$ denote the sum in $(1)$, then Maxima's Zeilberger function finds the recurrence $(c+2)f(c+2)=(c-b-a)f(c)+(b-a)f(c+1)$, and applying solve_rec to this returns false which implies that no hypergeometric solution exists. – ccorn Dec 27 '13 at 16:56
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Reference: Petkovšek, Wilf, Zeilberger (1997): $A=B$ – ccorn Dec 27 '13 at 17:13

Vandermonde-like sum

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