A polynomial that equals 1 / C(n+1, k) at k = 0, …, n

Idea. The weights $\binom{n+1}k$ in the alternating sum exactly cancel the given values.

The fact used. For a polynomial $Q$ of degree $<N$, $\sum_{k=0}^N(-1)^k\binom NkQ(k)=(-1)^N(\Delta^NQ)(0)=0$, because $\Delta$ lowers the degree by one at each application and $\deg P\le n<n+1=N$.

Apply it with $N=n+1$.
$$0=\sum_{k=0}^{n}(-1)^k\binom{n+1}k\cdot\frac1{\binom{n+1}k}+(-1)^{n+1}\binom{n+1}{n+1}P(n+1)=\sum_{k=0}^n(-1)^k+(-1)^{n+1}P(n+1).$$

Evaluate. $\sum_{k=0}^n(-1)^k$ equals $1$ if $n$ is even and $0$ if $n$ is odd. Hence
$$P(n+1)=(-1)^n\sum_{k=0}^n(-1)^k=\begin{cases}1&n\text{ even},\\0&n\text{ odd}.\end{cases}$$

Check. $n=1$: $P(0)=1$, $P(1)=\frac12$ gives $P(x)=1-\frac x2$ and $P(2)=0$ ✓. $n=2$: $P(0)=1$, $P(1)=\frac13$, $P(2)=\frac13$ gives $P(3)=1$ ✓.

Lesson. When a polynomial of degree $\le n$ is known at $n+1$ points, finite differences give its value at the next point with no interpolation formula.