23.16. (USA, 75). The polynomial $P(x)$ of degree $n$ satisfies the equalities $P(k)=k /(k+1)$ for $k=0,1, \ldots, n$. Find $P(n+1)$.
23.16. Note that there exists only one polynomial $P(x)$ satisfying the conditions of the problem, since if there were another polynomial $Q(x) \not \equiv P(x)$ with the same properties, then the polynomial $P(x)-Q(x)$ of degree not greater than $n$ would have at least $n+1$ roots. Since for the polynomial $R(x)=x+\frac{1}{(n-1-1)!}(0-x)(1-x) \ldots(n-x)$ the condition $R(-1)=0$ is satisfied, then by the theorem of Bezout this polynomial is divisible by $x+1$, i.e., $R(x)=S(x)(x+1)$, where $S(x)$ is a polynomial of degree $n$. Since $R(k)=k$ and $S(k)=k /(k-1)$ for $k=0,1, \ldots, n$, then $S(x)$ satisfies the conditions of the problem, and thus $P(x) \equiv S(x)$ and

$$
P(n+1)=\frac{R(n+1)}{n+2}=\frac{n+1+(-1)^{n+1}}{n+2}
$$