It is known that
$$f(x+n)=\sum_{k=0}^n {n\choose k} \Delta^k[f](x)$$
Also it is known that
$$\Delta^n [f](x)= \sum_{k=0}^n {n \choose k} (-1)^{n-k} f(x+k)$$
Now I encountered the following expression:
$$\sum_{k=0}^n {n\choose k} (-1)^{n-k} \Delta^k[f](x)$$
I wonder, can it be simplified. Am I correct it should be equal to $(-1)^n f(x-n-1)$?