Problem: Consider a polynomial (monic) $P$ and finite differences as $Q(x):= P\left(x+\tfrac{1}{2n}\right)-P(x)$ and so on, for any real $a$. Show that the sum $$\sum_{k=1}^n (-1)^k \dbinom{n}{k} P\left(x+\frac{1}{2n} \cdot k\right)=\left(\frac{1}{2n}\right)^d d!$$ where $d=\text{deg} P$
We know that the resulting $d^{\text{th}}$ difference is $\left(\frac{1}{2n}\right)^d d!$ but how to prove that the above expression is the same?
Thank you!
EDIT: Here is the relevant link to avoid any confusion of some sort.
