Example 8. If $\mathrm{P}(\mathrm{x})$ represents an $\mathrm{n}$-degree polynomial, and it is known that $\mathrm{P}(\mathrm{k})$ $=\frac{k}{1+k}(k=0,1,2,3, \cdots, n)$, try to find $\mathrm{P}(\mathrm{n}+\mathrm{1})$.
Construct a polynomial of degree $\mathrm{n}+1$:
$Q(x)=(x+1) P(x)-x$, according to the given conditions we have
$Q(\mathrm{k})=0 \quad(\mathrm{k}=0,1,2, \ldots, \mathrm{n})$,
and $Q(-1)=1$, hence we can obtain
$$
Q(x)=\frac{(-1)^{n+1}}{(x+1)!} x(x-1)(x-2) \cdots(x-n) \text {. }
$$

Thus,
$$
P(x)=\frac{(-1)^{n+1} \frac{1}{(n+1)!} x(x-1) \cdots(x-n)+x}{x+1} \text {. }
$$

Let $x=n+1$, we get
$$
\begin{aligned}
P(n+1) & =\frac{n+1+(-1)^{n+1}}{n+2} \\
& =\left\{\begin{array}{cc}
\frac{n}{n+2} ; & (n \text { is even) } \\
1 . & \text { ( } n \text { is odd) }
\end{array}\right.
\end{aligned}
$$