simplifying combinatorial sum

[Nico Schlömer]

Given natural numbers n, r, 0<=r<=n, I'm dealing with the expression

I have the strong suspicion that this can be simplified to a product of binomial coefficients. Is there any way that sympy can help me here?

[Aaron Meurer]

You can try using combsimp(). I think it has also received some improvements in master. 

You can also try using SymPy to evaluate the sum.

Aaron Meurer

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[Nico Schlömer]

Turns out the sum is always 1(!). I wasn't able to prove this with sympy though. This

```

import sympy

n = sympy.Symbol('n')

r = sympy.Symbol('r')

i = sympy.Symbol('i')

j = sympy.Symbol('j')

s = sympy.Sum(sympy.Sum(

    (-1)**(i+j)

    * sympy.binomial(n+r+1, i) * sympy.binomial(n-r, i)

    * sympy.binomial(n+r+1, j) * sympy.binomial(n-r, j)

    / sympy.binomial(2*n+1, i+j),

    (j, 0, n-r)), (i, 0, n-r))

```

isn't very cooperative with `doit()`.

On Thu, Nov 23, 2017 at 8:15 PM Aaron Meurer <asme...@gmail.com> wrote:

You can try using combsimp(). I think it has also received some improvements in master. 

You can also try using SymPy to evaluate the sum.

Aaron Meurer

On Thu, Nov 23, 2017 at 2:43 AM, Nico Schlömer <nico.sc...@gmail.com> wrote:

Given natural numbers n, r, 0<=r<=n, I'm dealing with the expression

I have the strong suspicion that this can be simplified to a product of binomial coefficients. Is there any way that sympy can help me here?

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