I am trying to find a closed form for the following double summation.
$\sum_{n=0}^{N} \sum_{m=0}^{N-n} \binom{n}{m}\binom{N-n}{m} x^n y^m$
It seems to be like Vandermonde's identity. Can anyone provide any suggestion?
Finding closed form for double summation binomial
2
$\begingroup$
combinatorics
summation
binomial-coefficients
closed-form
-
0Have you made any progress? What have you tried? – 2017-02-04
-
0Unfortunately, I could not make much progress. I tried to write as follow but it does not help $(1+y)^{N-n} = \sum_{m=0}^{N-n} \binom{N-n}{m} y^m$. – 2017-02-04
-
0This can trivially be transformed into $$=\sum_{n=0}^N x^n\sum_{m=n}^N\binom{n}{N-m}\binom{N-n}{N-m}y^{N-m} = \sum_{m=0}^N\sum_{n=0}^m \binom{n}{N-m}\binom{N-n}{N-m} x^n y^{N-m}$$ but I haven't been successful from there – 2017-02-04
-
0what makes you think this sum has a closed form? – 2017-02-04