9. let $P$ be a polynomial of degree $n$ such that

$$
P(k)=\frac{k}{k+1} \quad \text { for } \quad k=0,1,2, \ldots, n
$$

Find $P(n+1)$.

## Solution
Consider the polynomial $Q(x)=(x+1) P(x)-x$. $Q$ has degree $n+1$ and by assumption the $n+1$ zeros $k=0,1, \ldots, n$. There is therefore a constant $a$ with

$$
Q(x)=a x(x-1)(x-2) \cdots(x-n)
$$

In addition, $Q(-1)=((-1)+1) P(x)-(-1)=1$. Inserting $x=-1$ into the above formula then yields $1=(-1)^{n+1}(n+1)$ ! $\cdot a$, i.e. $a=(-1)^{n+1} /(n+1)$ ! From this now follows $Q(n+1)=(-1)^{n+1}$ and thus

$$
P(n+1)=\frac{Q(n+1)+(n+1)}{n+2}=\left\{\begin{array}{cl}
1 & n \text { odd } \\
\frac{n}{n+2} & n \text { even }
\end{array}\right.
$$