Given a positive integer $n$, my goal is to simplify the expression of the following function $$f(x,y)=\sum_{i=0}^n\binom{n}{i}\left((x-y)^{n+i}-(x+y)^{n+i}\right).$$ I know that with the binomial theorem, we have: $$(x-y)^{n+i}-(x+y)^{n+i}=\sum_{j=0}^{n+i}\left(\left(-1\right)^{n+i-j}-1\right)\binom{n+i}{j}x^jy^{n+i-j},$$ leading to $$f(x,y)=\sum_{i=0}^n\binom{n}{i}\left(\sum_{j=0}^{n+i}\left(\left(-1\right)^{n+i-j}-1\right)\binom{n+i}{j}x^jy^{n+i-j}\right).$$ $f$ is then the sum of $O(n^2)$ terms. When I develop the expression, for example with $n=3$, I obtain that $f$ is made of $x^0y^3$, $x^2y$, $xy^3$, $x^3y$, $x^0y^5$, $x^2y^3$, $x^4y$, $xy^5$, $x^3y^3$ ans $x^5y$. I would like to gather these terms as $$(x^2+x^3+x^4+x^5)y+(x^0+x+x^2+x^3)y^3+(x^0+x)y^5,$$ but I am not able to write such an explicit expression for $f$ with the correct coefficients.
So my question is, is it possible to write $f$ as $$f(x,y)=\sum_{...}^{...}\left(\sum_{...}^{...} ... x^{...}\right)y^{...}$$
Thank you!
Let
$f_{x}(y)=\sum_{i=0}^n\binom{n}{i}\left((x-y)^{n+i}-(x+y)^{n+i}\right) = \sum c_k(x) * y^k$.
But this is equivalent to writing $f$ as a taylor expansion in $y$. Thus,
$c_k(x) = \frac{1}{k!}\frac{\partial^k f_{x}(0)}{\partial y^k} = \sum_{i=0}^n\binom{n}{i}\left(\frac{(n+i-1)!}{(n+i-k-1)!}((-1)^k*x^{n+i-k}-x^{n+i-k})\right)$
It can be seen that $c_{k}(x)=0$ and $c_k(x)=\sum_{i=0}^n\binom{n}{i}\left(\frac{-2(n+i-1)!}{(n+i-k-1)!}x^{n+i}\right)$ when $k$ is even and odd respectively. This is how we can write $f$ in the way you required.