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 <asmeu...@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.schloe...@gmail.com>
> wrote:
>
>> Given natural numbers n, r, 0<=r<=n, I'm dealing with the expression
>>
>>
>> <https://lh3.googleusercontent.com/-hhIypvImNS4/WhaYAHKyE4I/AAAAAAABNUk/wBJbs4K02j0ftJtmWvD8uEKggygSn7VHACLcBGAs/s1600/sums.png>
>>
>>
>> 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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