Symmetry in function given by double sum I had to deal with this function:
$$ f_n(x_1,x_2)=(x_2-x_1)^{n-1}\sum_{m=0}^{n-1}\sum_{j=0}^{n-m-1}C(n,m,j)\left(\frac{x_2}{x_2-x_1}\right)^m\left(\frac{x_2(1-x_1)}{x_2-x_1}\right)^j $$
where 
$$C(n,m,j)=\frac{(-n+1)_m}{m!}\frac{(-n+m+1)_j(n)_j}{j!(m+2)_j}$$
or
$$C(n,m,j)=\frac{(-1)^{m+j}(n+j-1)!(m+1)}{j!(n-m-j-1)!(m+j+1)!}$$
(Here $(x)_j=x(x+1)\cdots (x+j-1)$ is the Pochhammer symbol or rising factorial).
I computed several cases and it seems that $f_n(x_1,x_2)$ is a symmetic function for every $n$, even though that is not apparent at first sight. Can one bring the function to another form that is explicitly symmetric?
I have tried playing with hypergeometric identites, but there are so many of them and I didn't get anywhere.
EDIT:
Playing with particular cases, it seems the function is given by
$$f_n(x_1,x_2)=\sum_{i=1}^{n}\sum_{j=1}^iA_{i,j}x_1^{n-1-i+j}x_2^{n-j}$$
and some of the coefficients are recognizable:
$$A_{i,1}=A_{i,i}={2n-i-1\choose n-1}{n-1\choose n-i}\frac{1}{(n-i+1)}$$
and
$$A_{n,j}={n-1\choose j-1}{n\choose j-1}\frac{1}{j}$$
 A: Let me change a bit your notation and put
$$
\eqalign{
  & g(x,y,n) = \left( {y - x} \right)^{\,n} \sum\limits_{0\, \le \,k\, \le \,n} {\sum\limits_{0\, \le \,j\, \le \,n - k} {
 D(n,k,j)\left( {{y \over {y - x}}} \right)^{\,k + j} \left( {1 - x} \right)^{\,j} } }  =   \cr 
  &  = \left( {y - x} \right)^{\,n} \sum\limits_{0\, \le \,k\, \le \,n} {\sum\limits_{k\, \le \,j + k\, \le \,n} {\sum\limits_{0\, \le \,i\,\left( { \le \,j} \right)} {
 \left( { - 1} \right)^{\,i} \binom{j+k-k}{i} D(n,k,j){{y^{\,k + j} \,x^{\,i} } \over {\left( {y - x} \right)^{\,k + j} }}} } }  =   \cr 
  &  = \left( {y - x} \right)^{\,n} \sum\limits_{0\, \le \,k\, \le \,n} {\sum\limits_{k\, \le \,l\, \le \,n} {\sum\limits_{0\, \le \,i\,\left( { \le \,j} \right)} {
 \left( { - 1} \right)^{\,i} \binom{ l - k }{i} D(n,k,l){{y^{\,l} \,x^{\,i} } \over {\left( {y - x} \right)^{\,l} }}} } }  \cr} 
$$
where 
$$
\eqalign{
  & D(n,k,j) = {{\left( { - n} \right)^{\,\overline {\,k\,} } } \over {k!}}{{\left( { - \left( {n - k} \right)} \right)^{\,\overline {\,j\,} } \left( {n + 1} \right)^{\,\overline {\,j\,} } } \over {j!\left( {k + 2} \right)^{\,\overline {\,j\,} } }} =   \cr 
  &  = {{\left( { - 1} \right)^{\,k} n^{\,\underline {\,k\,} } } \over {k!}}{{\left( { - 1} \right)^{\,j} \left( {n - k} \right)^{\,\underline {\,j\,} } \left( {n + 1} \right)^{\,\overline {\,j\,} } } \over {j!\left( {k + 2} \right)^{\,\overline {\,j\,} } }} =   \cr 
  &  = \left( { - 1} \right)^{\,k + j} \left( \matrix{
  n \cr 
  k \cr}  \right)\left( \matrix{
  n - k \cr 
  n - \left( {k + j} \right) \cr}  \right){{\left( {k + 2 + \left( {n - k} \right) - 1} \right)^{\,\overline {\,j\,} } } \over {\left( {k + 2} \right)^{\,\overline {\,j\,} } }} =   \cr 
  &  = \left( { - 1} \right)^{\,k + j} \left( \matrix{
  n \cr 
  k \cr}  \right)\left( \matrix{
  n - k \cr 
  n - \left( {k + j} \right) \cr}  \right){{\left( {k + 2 + \left( {n - k} \right) - 1} \right)^{\,\overline {\,j\,} } } \over {\left( {k + 2} \right)^{\,\overline {\,j\,} } }} =   \cr 
  &  = \left( { - 1} \right)^{\,k + j} \left( \matrix{
  n \cr 
  k \cr}  \right)\left( \matrix{
  n - k \cr 
  n - \left( {k + j} \right) \cr}  \right){{\left( {k + j + 2} \right)^{\,\overline {\,\left( {n - k} \right) - 1\,} } } \over {\left( {k + 2} \right)^{\,\overline {\,\left( {n - k} \right) - 1\,} } }} \cr} 
$$
We can then replace $k+j$ with $l$
$$
D(n,k,l) = \left( { - 1} \right)^{\,l} \binom{n}{k} \binom{n-k}{n-l}
{{\left( {l + 2} \right)^{\,\overline {\,\left( {n - k} \right) - 1\,} } } \over {\left( {k + 2} \right)^{\,\overline {\,\left( {n - k} \right) - 1\,} } }}
$$
Note that the bounds
$$
0\le k \le n  \quad
k\le l \le n
$$
are implicit in the binomials and can therefore be virtually omitted 
in the sum for $g(x,y,n)$
Thereafter, some further algebraic manipulation on 
the above expression leads to:
$$
\eqalign{
  & D(n,k,l) = \left( { - 1} \right)^{\,l} \left( \matrix{
  n \cr 
  k \cr}  \right)\left( \matrix{
  n - k \cr 
  n - l \cr}  \right){{\left( {l + 2} \right)^{\,\overline {\,\left( {n - k} \right) - 1\,} } } \over {\left( {k + 2} \right)^{\,\overline {\,\left( {n - k} \right) - 1\,} } }} =   \cr 
  &  = \left( { - 1} \right)^{\,l} {{n^{\,\underline {\,\left( {n - k} \right)\,} } } \over {\left( {n - k} \right)!}}{{\left( {n - k} \right)!} \over {\left( {n - l} \right)!
 \left( {l - k} \right)!}}{{\left( {n - k + l} \right)^{\,\underline {\,\left( {n - k} \right) - 1\,} } } \over {n^{\,\underline {\,\left( {n - k} \right) - 1\,} } }} =   \cr 
  &  = \left( { - 1} \right)^{\,l} {{n^{\,\underline {\,\left( {n - k} \right)\,} } } \over {\left( {n - l} \right)!\left( {l - k} \right)!}}{{
\left( {n - k + l} \right)^{\,\underline {\,\left( {n - k} \right) - 1\,} } } \over {n^{\,\underline {\,\left( {n - k} \right)\,} } k^{\,\underline {\, - 1\,} } }} =   \cr 
  &  = \left( { - 1} \right)^{\,l} {{\left( {n + l - k} \right)^{\,\underline {\,\left( {n - k} \right)\,} } } \over {\left( {n - l} \right)!
\left( {l - k} \right)!}}{{l^{\,\underline {\, - 1\,} } } \over {k^{\,\underline {\, - 1\,} } }} =   \cr 
  &  = \left( { - 1} \right)^{\,l} \left( \matrix{
  n + l - k \cr 
  n - k \cr}  \right)\left( \matrix{
  n - k \cr 
  l - k \cr}  \right){{k + 1} \over {l + 1}} =   \cr 
  &  = \left( { - 1} \right)^{\,l} \left( \matrix{
  n + l - k \cr 
  l - k \cr}  \right)\left( \matrix{
  n \cr 
  n - l \cr}  \right){{k + 1} \over {l + 1}} \cr} 
$$
Adding the multiplying binomial, the complete factor becomes
$$
\eqalign{
  & F(n,k,l,i) = \left( { - 1} \right)^{\,i} \binom{l-k}{i} D(n,k,l) =   \cr 
  &  = \left( { - 1} \right)^{\,i} \left( { - 1} \right)^{\,l} \binom{l-k}{i}
 \binom{n+l-k}{l-k} \binom{n}{n-l}
{{k + 1} \over {l + 1}} =   \cr 
  &  = \left( { - 1} \right)^{\,l - i} \binom{n+l-k}{l-k}  \binom{l-k}{l-k-i}  \binom{n}{n-l}   {{k + 1} \over {l + 1}} =   \cr 
  &  = \binom{n+l-k}{l-i-k} \binom{k+1}{k}
 \left( { - 1} \right)^{\,l - i} \binom{n+i}{i} \binom{n}{n-l}{1 \over {l + 1}} \cr} 
$$
Note that we have manouvred in such a way as to group the
terms in $k$. We can therefore sum $F(n,k,l,i)$ over $k$
using the "double convolution formula" for binomials.
Let's remember that the bounds on the sum can be waived
thus allowing to carry on the convolution without 
caring to check them.
So we get
$$
\eqalign{
  & G(n,l,i) = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,n} \right)} {F(n,k,l,i)}  =   \cr 
  & \left( { - 1} \right)^{\,l - i} \binom{n+i}{i} \binom{n}{n-l}
  {1 \over {l + 1}}\sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,l - i} \right)} {
  \binom{n + l - k}{  l - i - k} \binom{  k + 1}{  k } }  =   \cr 
  &  = \left( { - 1} \right)^{\,l - i} \binom{  n + i}{  i }\binom{  n}{  n - l } {1 \over {l + 1}}\binom{  n + l + 2 }{  l - i }  =   \cr 
  &  = \left( { - 1} \right)^{\,l - i} \binom{  n + i }{  i}\binom{  n }{  l } {1 \over {l + 1}}\binom{  n + l + 2 }{  l - i }   =   \cr 
  &  = \left( { - 1} \right)^{\,l - i} {1 \over {n + 1}}\binom{  n + i }{  i } \binom{  n + 1 }{  l + 1} \binom{  n + l + 2}{  l - i}  \cr} 
$$
I suppose that from here you can conclude by yourself.
