Proving that $f$ verifies $f(1-x)=-f(x)$ I'm trying to show an identity verified by a function. I have this function which is defined for all real numbers as
$ f(x) = \sum_{k=0}^{p-1} \frac{(x+k)^{2m-1}}{(-1)^{p+k} (p-1-k)!(p+k)!}$
I would like to show that $f(1-x) = -f(x)$.
First thing that tried:
$
\begin{align*} 
f(1-x)
&= \sum_{k=0}^{p-1} \frac{(1-x+k)^{2m-1}}{(-1)^{p+k} (p-1-k)!(p+k)!} \\
&= \sum_{k=0}^{p-1} \frac{(-1)^{2m-1}(x-1-k)^{2m-1}}{(-1)^{p+k} (p-1-k)!(p+k)!} 
\quad \text{factoring the numerator by }(-1)^{2m-1}\\
&= -\sum_{k=0}^{p-1} \frac{(x-1-k)^{2m-1}}{(-1)^{p+k} (p-1-k)!(p+k)!}
\quad \text{reducing }(-1)^{2m-1}=-1\\
&= -\sum_{i=-p}^{-1} \frac{(x+i)^{2m-1}}{(-1)^{p-i-1} (p+i)!(p-i-1)!}
\quad \text{doing a change of variable } i = -k-1 \\
\end{align*}
$
I'm kind of stuck at this point and I don't see how to move forward with it. I think I'm not going to the optimal direction to prove that relation.
Second thing that I am investigating:
Since $f(x)$ is a polynomial of degree $2m-1$, I could write it as a finite power series centered at $0$. So with Taylor's theorem I would have
$f(x) = \sum_{i=0}^{2m-1}  \frac{f^{(i)}(0) }{i!} x^{i}$.
and
$
\begin{align*} 
f(1-x) 
&= \sum_{i=0}^{2m-1}  \frac{ f^{(i)}(0) }{i!} (1-x)^{i} \\
&= \sum_{i=0}^{2m-1}  \frac{ f^{(i)}(0) }{i!} \Bigg( \sum_{j=0}^{i} \binom{i}{j} (-x)^{j} \Bigg)
\quad \text{binomial expansion applied to } (1-x)^{i} \\
&= \sum_{j=0}^{2m-1}  \Bigg((-1)^{j}\sum_{i=j}^{2m-1} \binom{i}{j} \frac{ f^{(i)}(0) }{i!} \Bigg) x^{j}
\quad \text{changing the summation order} \sum_{i=0}^{2m-1}\sum_{j=0}^{i} \text{ to }   \sum_{j=0}^{2m-1}\sum_{i=j}^{2m-1}  \\
\end{align*}
$
So showing that $f(1-x) = -f(x)$ is equivalent to showing that
$(-1)^{j}\sum_{i=j}^{2m-1} \binom{i}{j} \frac{ f^{(i)}(0) }{i!} = - \frac{f^{(j)}(0) }{j!} $.
I think this is the more elegant way to go but I don't know if it is going to make things worse.
Any feedback ?
EDIT : clarification of the parameters m and p :
I forgot to mention hypothesis on the parameters $m$ and $p$. So we consider that $ 1 \leq m < p$

We can rewrite the sum as
$$
\eqalign{
  & f\left( x \right) = \sum\limits_{k = 0}^{p - 1} {{{\left( {x + k} \right)^{\,2m - 1} }
 \over {\left( { - 1} \right)^{\,p + k} \left( {p - 1 - k} \right)!\left( {p + k} \right)!}}}  =   \cr 
  &  = \sum\limits_{k = 0}^{p - 1} {\left( { - 1} \right)^{\,p + k} {{\left( {x + k} \right)^{\,2m - 1} }
 \over {\left( {p - 1 - k} \right)!\left( {p + k} \right)!}}} \quad  \Rightarrow   \cr 
  &  \Rightarrow \quad g\left( x \right) = \left( {2p - 1} \right)!f\left( x \right) =   \cr 
  &  = \sum\limits_{k = 0}^{p - 1} {\left( { - 1} \right)^{\,p + k}
 \left( \matrix{ 2p - 1 \cr p + k \cr}  \right)\left( {x + k} \right)^{\,2m - 1} }  \cr} 
$$
Then for the sum $g(1-x)+g(x)$ we get
$$
\eqalign{
  & g\left( {1 - x} \right) + g(x) =   \cr 
&  = \sum\limits_{k = 0}^{p - 1} {\left( { - 1} \right)^{\,p + k}
 \left( \matrix{2p - 1 \cr p + k \cr}  \right)\left( {1 - x + k} \right)^{\,2m - 1} }  +   \cr 
  &  + \sum\limits_{k = 0}^{p - 1} {\left( { - 1} \right)^{\,p + k}
 \left( \matrix{2p - 1 \cr p + k \cr}  \right)\left( {x + k} \right)^{\,2m - 1} }  =   \cr 
  &  = \left( { - 1} \right)^{\,2m - 1} \sum\limits_{k = 0}^{p - 1}
 {\left( { - 1} \right)^{\,p + k} \left( \matrix{ 2p - 1 \cr  p + k \cr}  \right)
\left( {x - k - 1} \right)^{\,2m - 1} }  +   \cr 
  &  + \sum\limits_{k = 0}^{p - 1} {\left( { - 1} \right)^{\,p + k}
 \left( \matrix{ 2p - 1 \cr   p + k \cr}  \right)\left( {x + k} \right)^{\,2m - 1} }  =   \cr 
  &  =  - \left( { - 1} \right)^{\,2m - 1} \sum\limits_{k = 0}^{p - 1} {\left( { - 1} \right)^{\,p - 1 - k}
 \left( \matrix{2p - 1 \cr p - 1 - k \cr}  \right)\left( {x - p + p - 1 - k} \right)^{\,2m - 1} }  +   \cr 
  &  + \sum\limits_{k = 0}^{p - 1} {\left( { - 1} \right)^{\,p + k}
 \left( \matrix{2p - 1 \cr p + k \cr}  \right)\left( {x - p + p + k} \right)^{\,2m - 1} }  =   \cr 
  &  = \sum\limits_{k = 0}^{p - 1} {\left( { - 1} \right)^{\,k}
 \left( \matrix{2p - 1 \cr k \cr}  \right)\left( {x - p + k} \right)^{\,2m - 1} }  +   \cr 
  &  + \sum\limits_{k = p}^{2p - 1} {\left( { - 1} \right)^{\,k}
 \left( \matrix{2p - 1 \cr k \cr}  \right)\left( {x - p + k} \right)^{\,2m - 1} }  =   \cr 
  &  = \sum\limits_{k = 0}^{2p - 1} {\left( { - 1} \right)^{\,k}
 \left( \matrix{2p - 1 \cr k \cr}  \right)\left( {x - p + k} \right)^{\,2m - 1} }  =   \cr 
  &  = \left( { - 1} \right)^{\,2p - 1} \Delta ^{\,2p - 1} \left( {x - p} \right)^{\,2m - 1}  = 0 
\quad \;\left| {\,1 \le m < p} \right. \cr}
$$
and that is the $2p-1$-th finite difference of a polynomial of degree $2m-1$
which is null if $ m < p$
