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I'd like to get Mathematica (11) to do basic sums, such as

Sum[Binomial[a, b - k] Binomial[c, d + k], {k, -d, b}].

This is the Chu-Vandermonde sum, which equals ${a+c\choose b+d}$, at least for integer $a,c$.

Mathematica instead evaluates the sum to

Binomial[a, b + d] Hypergeometric2F1[-c, -b - d, 1 + a - b - d, 1],

which is true, but not as useful.

My guess was, that I needed to somehow specify that $a,b,c,d$ were positive integers, or something, so I entered the following (BTW, is there an easier way to do this?)

$Assumptions =  a ∈ Integers && b ∈ Integers && c ∈ Integers && d ∈ Integers
                && a > 0 && b > 0 && c > 0 && d > 0

An evaluated the sum again, this time with a FullSimplify. However it didn't work, and I still got the hypergeometric representation.

Is this because Mathematica doesn't know the formula? Or is there some assumption I didn't tell it? The problem is not just with Vandermonde, but with many similar sums, which I think Mathematica should be able to put on a nice closed form, but for some reason it doesn't.

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Add to the the assumptions that 1 + a > b + d

assume = Element[{a, b, c, d}, Integers] &&
   a > 0 && b > 0 && c > 0 && d > 0 && 1 + a > b + d;

expr = Sum[Binomial[a, b - k] Binomial[c, d + k], {k, -d, b}]

(*  Binomial[a, b + d] Hypergeometric2F1[-c, -b - d, 1 + a - b - d, 1]  *)

The additional assumption comes from requiring that the third argument of the Hypergeometric2F1 be a positive integer.

To eliminate the Hypergeometric2F1 use FunctionExpand

expr2 = Assuming[assume, expr //
    FunctionExpand // FullSimplify]

(*  (a + c)!/((a - b + c - d)! (b + d)!)  *)

Verifying that given the assumptions that this is equivalent to the Binomial

Assuming[assume,
 expr2 == Binomial[a + c, b + d] // FullSimplify]

(*  True  *)

EDIT: Using a replacement rule to simplify

rule = (Times[Factorial[x_], Power[Factorial[y_], -1],
       Power[Factorial[z_], -1]] /; z == x-y) :> Binomial[x, z];

expr3=expr2/.rule

(*  Binomial[a+c, b+d]  *)
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  • $\begingroup$ Thank you! That answered nearly all of my questions, including with regards to the assumption syntax :-) But why do we require that the third argument is a positive integer? mathworld.wolfram.com/GausssHypergeometricTheorem.html only seems to require $1 + a + c > 0$? Also, I assume there is no way to make Mathematica take the last step of reassembling the binomial? $\endgroup$ – Thomas Ahle Dec 20 '16 at 19:11
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    $\begingroup$ @ThomasAhle - since you assumed that all variables are integers then 1+a-b-d is an integer. If it is not a positive integer then there will be a discontinuity unless either c or b+d has a smaller magnitude than the magnitude of 1+a-b-d. Rather than work out all the details I simplified the problem to 1 + a > b + d. To rewrite as a Binomial you would need to Simplify or FullSimplify with a custom ComplexityFunction $\endgroup$ – Bob Hanlon Dec 20 '16 at 19:39
  • $\begingroup$ @Thomas, recall the definition of the Gaussian hypergeometric function, whose series at $0$ includes a Pochhammer symbol in the denominator. Since $(-j)_k=0,\,k>j>0$, hypergeometric functions do not admit these numbers as "denominator parameters". $\endgroup$ – J. M. is away Dec 21 '16 at 1:12

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