Questions on a self-made theorem about polynomials I recently came up with this theorem:

For any complex polynomial $P$ degree $n$:
$$ \sum\limits_{k=0}^{n+1}(-1)^k\binom{n+1}{k}P(a+kb) = 0\quad \forall a,b \in\mathbb{C}$$

Basically, if $P$ is quadratic, $P(a) - 3P(a+b) + 3P(a+2b) - P(a+3b) = 0$ (inputs of $P$ are consecutive terms of any arithmetic sequence). This can be generalized to any other degrees.

*

*Has this been discovered? If yes, what's the formal name for this phenomenon?

*Is it significant/Are there important consequences of this being true?

*Can this be generalized to non-polynomials?

 A: In brief: this is well-known, but definitely important.
It's easiest to write this in terms of the finite difference operator $\Delta$: $\Delta P(x)=P(x+1)-P(x)$.  You use $P(x+b)$ instead of $P(x+1)$, but it's easy to see that these two things are equivalent; to keep things consistent with your notation, I'll write $\Delta_b$ for your operator.
The most important feature of the $\Delta_b$ operator is how it affects the degree of a polynomial:
Theorem: for any nonconstant polynomial $P(x)$, the degree of $\Delta_b P(x)$ is one less than the degree of $P(x)$.
Proof outline: Note that the degree of $\Delta_b P(x)$ is no greater than the degree of $P(x)$.  Now, write $P(x) = a_dx^d+Q(x)$, where $Q(x)$ is a polynomial of degree $d-1$ or less.  Then $P(x+b) =a_d(x+b)^d+Q(x+b)$, so $\Delta_b P(x) = a_d\left((x+b)^d-x^d\right)+\Delta_b Q(x)$; by the binomial theorem $(x+b)^d=x^d+{d\choose 1}bx^{d-1}+\ldots$, so $(x+b)^d-x^d={d\choose 1}bx^{d-1}+\ldots$ is a polynomial of degree at most $d-1$, and thus $\Delta_bP(x)$ is the sum of two polynomials of degree at most $d-1$ (namely, $a_d\left((x+b)^d-x^d\right)$ and $\Delta_b Q(x)$), so it's of degree at most $d-1$ itself.
(It's slightly more challenging to prove that the degree of $\Delta_bP(x)$ is exactly $d-1$ when $b \neq 0$, but this can also be shown.)
Why does this matter? Because it can be shown by induction that your sum is exactly the result of applying the $\Delta_b$ operator $d+1$ times, where $d$ is the degree of the polynomial; since each application of $\Delta_b$ reduces the degree by one, then $(\Delta_b)^dP(x)$ is a polynomial of degree zero — a constant — and thus $(\Delta_b)^{d+1}P(x)$ will be identically zero.  This is exactly your identity.
Now, you may know that the derivative of a polynomial of degree $d$ is also a polynomial of degree $d-1$.  It turns out that this isn't a coincidence; $\Delta$ is very similar to a derivative in many ways, with the Newton polynomials ${x\choose d}=\frac1{d!}x(x-1)(x-2)\cdots(x-d)$ playing the role of the monomial $x^d$ with respect to the derivative.  For more details, I suggest starting with Wikipedia's page on finite difference calculus.
In fact, we can also prove the converse (and this answers the question about generalizing to non-polynomials in the negative).  I'll work in terms of $\Delta$, rather than $\Delta_b$, but again all the results generalize readily.
Note that $\Delta^n P(x)$ only depends on the values of $P(x+i)$ for $i$ an integer between $0$ and $n$; thus, a function can take arbitrary values for $0\lt x\lt1$ and still satisfy the identity; we can't say much about general points. However, it does constrain the values at integers:
Theorem: suppose that $\Delta^{d+1}f(x)\equiv 0$ identically. Then there exists a polynomial $P(x)$ of degree $d$ such that $f(n)=P(n)$ for all integers $n$.
The proof works by induction. For simplicity's sake, I'll consider all functions as being on $\mathbb{Z}$ now, and not consider non-integer values at all.  Note first of all that if $\Delta f(x)=g(x)$, then $f(n)=f(0)+\sum_{i=0}^{n-1}g(i)$.  (Proof by induction: the case $n=1$ is true by definition, since $g(0)=\Delta f(0)=f(1)-f(0)$ implies that $f(1)=f(0)+g(0)$. Now, assuming it's true for $n=k$, at $n=k+1$ we have $f(k+1)=f(k)+g(k)$ $=f(0)+\sum_{i=0}^{k-1}g(i)+g(k)$ $=f(0)+\sum_{i=0}^kg(i)$.)  In particular, if $\Delta f(x)\equiv 0$ identically, then $f(n)=f(0)$ for all integers $n$; $f()$ is constant on $\mathbb{Z}$.
This gives us the base case for our induction; to induct we just need to show that if $\Delta f(x)$ is a polynomial of degree $d$, then $f(x)$ is polynomial of degree $d+1$.  But suppose for concreteness that $\Delta f(x)=P(x)=\sum_{i=0}^da_ix^i$.  Then $f(n)=f(0)+\sum_{k=0}^{n-1}P(k)$ $=f(0)+\sum_{k=0}^{n-1}\left(\sum_{i=0}^da_ik^i\right)$ $=f(0)+\sum_{i=0}^da_i\left(\sum_{k=0}^{n-1}k^i\right)$.  Now, for each $i$ the sum $\sum_{k=0}^{n-1}k^i$ in parentheses in this last expression is known to be a polynomial of degree $i+1$ (see e.g. https://en.wikipedia.org/wiki/Faulhaber%27s_formula ), so the whole expression is a polynomial of degree $d+1$, as was to be proved.
A: What you wrote is the finite difference operator of order $n+1$, acting on $P$,
$$ \sum\limits_{k=0}^{n+1}(-1)^k\binom{n+1}{k}P(a+kb) =\Delta^{n+1} P(a+kb).$$
Notice that by linearity it suffices for the property to hold for all monomials $k^m, m\le n$ 
and it is easily explained by the fact that the first order difference of a polynomial is a polynomial of degree one less.
$$(k+1)^m-k^m=k^m+mk^{m-1}+\cdots-k^m.$$

Illustration ($n=3$):
$$\Delta^4 k^m=((4^m-3^m)-(3^m-2^m))-((3^m-2^m)-(2^m-1^m))
\\-((3^m-2^m)-(2^m-1^m))-((2^4-1^m)-(1^4-0^m))
\\=4^m-4\cdot3^m+6\cdot2^m-4\cdot1^m+0^m.$$
and
$$\begin{matrix}
1&&1&&1&&1&&1
\\&0&&0&&0&&0
\\&&0&&0&&0
\\&&&0&&0
\\&&&&0
\end{matrix}$$
$$\begin{matrix}
0&&1&&2&&3&&4
\\&1&&1&&1&&1
\\&&0&&0&&0
\\&&&0&&0
\\&&&&0
\end{matrix}$$
$$\begin{matrix}
0&&1&&4&&9&&16
\\&1&&3&&5&&7
\\&&2&&2&&2
\\&&&0&&0
\\&&&&0
\end{matrix}$$
$$\begin{matrix}
0&&1&&8&&27&&64
\\&1&&7&&19&&37
\\&&6&&12&&18
\\&&&6&&6
\\&&&&0
\end{matrix}$$

Final remark:
On can show that $\Delta_{n+1}k^{n+1}=(-1)^nn!$, so that for a polynomial of degree $n+1$ the sum is
$$(-1)^nn!p_{n+1}b^{n+1},$$ independently of $a$.
A: This result is known and was demonstrated by a student named Ruiz.
Here's the reference :
Sebastián Martín Ruiz, An Algebraic Identity Leading to Wilson's Theorem, 
The Mathematical Gazette, 80 (489)  579-582 (Nov. 1996).
You accesss to it at JSTOR : http://www.jstor.org/stable/3618534 or on the arXiv: arXiv:math/0406086.
A: This is not a new result. What you wrote is the formula for the $n$-th order forward difference and can be found even on Wikipedia.
I will show a representation formula which is valid for $C^n(\mathbb R)$ functions, not just polynomials. The same can be found also here.
The key point is that finite differences and derivatives commute: $D\Delta_h=\Delta_hD$.
For $f\in C^1(\mathbb R)$ you can compute
$$
\frac1h\Delta_h[f](x) = \frac{f(x+h)-f(x)}h
= \frac1h \int_0^h D[f](x+x_1) \,dx_1
$$
For $f\in C^n(\mathbb R)$, iterating the above formula, you get
$$
\begin{split}
\frac1{h^n}\Delta_h^n[f](x)
&= \frac{\frac1{h^{n-1}}\Delta_h^{n-1}[f](x+h)
 - \frac1{h^{n-1}}\Delta_h^{n-1}[f](x)} {h} \\
&= \frac1h \int_0^h \frac1{h^{n-1}}D\Delta_h^{n-1}[f](x+x_1) \,dx_1 \\
&= \frac1{h^2} \int_0^h \int_0^h
 \frac1{h^{n-1}} D^2\Delta_h^{n-2}[f](x+x_1+x_2) \,dx_1dx_2 \\
&= \,\cdots \\
&= \frac1{h^n} \int_0^h\dotsi\int_0^h D^n[f](x+x_1+\dotsb+x_n) \,dx_1\dotsm dx_n .
\end{split}
$$
The last integral is a weighted average of $D^n[f]$ over the segment $[x,x+nh]$. More precisely, let $X_1,\dotsc,X_n\sim\mathrm{Uniform}(0,h)$ be i.i.d. and $S=X_1+\dotsb+X_n$. Then the previous expression can be viewed as
$$
\frac1{h^n}\Delta_h^n[f](x)
= \mathbb{E}\left[D^n[f](S)\right]
$$
The mean value theorem for integrals applies and tells us that there exists $x^*\in(x,x+nh)$ such that
$$
\frac1{h^n}\Delta_h^n[f](x) = D^n[f](x^*).
$$
Regarding your proposition, it follows trivially from the fact that if $P$ is a polynomial of degree $n$, then $D^{n+1}[P]=0$ identically, hence also $\Delta_h^{n+1}[P]=0$.
I can recommend reading at least Finite difference, Indefinite sum and Newton polynomial to learn about related material.
