23.17. (Jury, 81). The polynomial $P(x)$ of degree $n$ satisfies the equalities $P(k)=1 / C_{n+1}^{k}$ for $k=0,1, \ldots, n$. Find $P(n+1)$.
23.17. Using the Lagrange interpolation formula, we obtain (below it is implied that $1 \leqslant i \leqslant n$)

$$
\begin{aligned}
& P(x)=\sum_{k=0}^{n} \frac{1}{C_{n+1}^{k}} \prod_{i \neq k} \frac{(x-i)}{(k-i)}=\sum_{k=0}^{n} \frac{\prod_{i \neq k}(x-i)}{\left.C_{n+1}^{k}(-1)^{n-k}(n-k)!k\right)}= \\
& =\sum_{k=0}^{n}(-1)^{n-k} \frac{(n+1-k)}{(n+1)!} \prod_{i \neq k}(x-i), \\
& \begin{aligned}
& P(n+1)=\sum_{k=0}^{n}(-1)^{n-k} \frac{(n+1-k)}{(n+1)!} \prod_{i \neq k}(n+1-i)= \\
&= \sum_{k=0}^{n} \frac{(-1)^{n-k}}{(n+1)!} \prod_{i=0}^{n}(n+1-i)=\sum_{k=0}^{n}(-1)^{n-k}
\end{aligned}
\end{aligned}
$$

Therefore, $P(n+1)=0$ if $n$ is odd, and $P(n+1)=1$ if $n$ is even. 340