# Prove or disprove that if $f(x)=\sum_{k=0}^na_kx^k$ then $\sum_{k=0}^{n+1}(-1)^{n+1-k}\binom{n+1}kf(x+k)=0$

I had the following problem:
Let $f(x)=ax^2+bx+c$. Find $f(x+3)-3f(x+2)+3f(x+1)-f(x)$.
It is very simple to solve this and the result is $0$. But, I noticed that coefficients of second equality $(1,-3,3,-1)$ is the same as coefficients of $(x-1)^3$. So, I came with the following hypothesis:

Let $f(x)=\sum_{k=0}^na_kx^k$ for all $x\in\mathbb{R}$ and some $n\in\mathbb{N},a_0,a_1,\dots,a_n\in\mathbb{R}$. Then $$\sum_{k=0}^{n+1}(-1)^{n+1-k}\binom{n+1}kf(x+k)=0$$

I tried to find some way to prove or disprove it. I firstly tried induction. It is true for $n=\{1,2,3\}$, so I supposed it is true for some $m\in\mathbb{N}$. Then I realized that it is not easy to prove it for $m+1$ because the coefficients isn't same. Also, it is very different for odd and even $n$.
My second attempt is to try to group terms in some way. So, I have tried to find $$\sum_{k=0}^{n}(-1)^{n-k}\binom nk$$ It is easy to prove that for any odd $n$ it is equal to $0$. Then I proved that for even $n$ we can group terms so we can reduce it to the first case (odd $n$), so for all $n\in\mathbb{N}$ it is equal to $0$.
Now, problem is $f(x+k)$. After many attempts, I couldn't find any way to group terms and reduce it to $0$.
What is the easiest way to prove or disprove it?

One of the reasons why I've written out m.se answer #1379518 in all that detail was my expectation that it would come up over and over. Well, this is such a case. Fix $x \in \mathbb{R}$. In the notations of this answer, the sequence $\left(f\left(x+0\right),f\left(x+1\right),\cdots,f\left(x+n+1\right)\right)$ is $n$-polynomial (since $f\left(x+t\right)$ is a polynomial in $t$ of degree $\leq n$), and thus Theorem 3 in my post (applied to $\mathbf{a} = \left(f\left(x+0\right),f\left(x+1\right),\cdots,f\left(x+n+1\right)\right)$ and $k=n$) shows that the sequence $\Delta^{n+1}\left(f\left(x+0\right),f\left(x+1\right),\cdots,f\left(x+n+1\right)\right)$ consists solely of zeroes. But Theorem 2 in the same post (applied to $n+2$, $\left(f\left(x+0\right),f\left(x+1\right),\cdots,f\left(x+n+1\right)\right)$, $n+1$, and $1$ instead of $n$, $\left(a_1,a_2,\ldots,a_n\right)$, $k$ and $p$) shows that the $1$-st entry of this sequence is $\sum\limits_{i=0}^{k+1}\left(-1\right)^i\dbinom{n+1}{i}f\left(x+i\right)$. Thus, $\sum\limits_{i=0}^{k+1}\left(-1\right)^i\dbinom{n+1}{i}f\left(x+i\right) = 0$. Multiply this equality by $\left(-1\right)^{n+1}$, and you obtain your claim (because $\left(-1\right)^{n+1} \cdot \left(-1\right)^i = \left(-1\right)^{n+1-i}$).

The claim holds if $$\sum_{k=0}^{n+1} {n+1\choose k} (-1)^k (x+k)^q = 0$$ where $0\le q\le n.$

Introduce $$(x+k)^q = \frac{q!}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{q+1}} \exp((x+k)z) \; dz.$$

This yields for the sum $$\frac{q!}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{q+1}} \exp(xz) \sum_{k=0}^{n+1} {n+1\choose k} (-1)^k \exp(kz) \; dz \\ = \frac{q!}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{q+1}} \exp(xz) (1-\exp(z))^{n+1}.$$

This is $$q! [z^q] \exp(kz) (1-\exp(z))^{n+1}.$$

Note however that $$1-\exp(z) = -z - \frac{1}{2} z^2 -\frac{1}{6} z^3 - \cdots$$

so that $(1-\exp(z))^{n+1}$ starts at $z^{n+1}.$ Hence the coefficient on $z^q$ is zero because we assumed that $0\le q\le n.$

Algebraic methode :Set $E=\Bbb R_n[X]=\{P\in \Bbb R[X] / \ \deg P\leq n \}$ and $\varphi:E\to E$ defined by $\varphi(P)=P(X+1)$ , then $\varphi$ is alinear map and $\varphi^k(P)=P(X+k)$. Now use the Cayley-Hamilton's Theorem.