Consider the following binomial identity: $$ \sum_{k=0}^n(-1)^k\binom{n}{k}g(k)=0 $$ for every polynomial $g(k)$ with degree less than $n$.
My Proof Every polynomial $g(k)$ of degree $t$ can be represented in the following form $$ g(k) = \sum_{l=0}^tc_l(k)_l, $$ where $$ (k)_l=k(k-1)\ldots(k-l+1), $$ ans $c_l$ are some coefficients.
For every $l Question
Do you know some other proofs of this identity? I'm most interested in combinatorial proof.
Let $n$ a positive integer and $E$ the real vector space of polynomials with degree $ Consider the linear map $\varphi:E\to E,P(X)\mapsto P(X)-P(X+1)$. For every nonzero $P(X)\in E$, we observe that $\deg(\varphi(P(X))<\deg(P(X))$. This leads to $\varphi^n=0$ (of course $\varphi^k$ means $\varphi\circ\cdots\circ\varphi$ with $k$ times $\varphi$). Now consider the map $\psi:E\to E,P(X)\mapsto P(X+1)$, so that $\varphi=id_E-\psi$. Since $\psi$ and $id_E$ commute, we can apply Newton's binomial formula and get : $$\sum_{k=0}^n{n\choose k}(-1)^k\psi^k=\left(id_E-\psi\right)^n=\varphi^n=0$$ So, for any $P(X)\in E$ : $$\sum_{k=0}^n{n\choose k}(-1)^kP(X+k)=0$$Evaluating in $0$, we finally get : $$\sum_{k=0}^n{n\choose k}(-1)^kP(k)=0$$