13. (ROM) Let $P$ be a polynomial of degree $n$ satisfying  
$$ P(k)=\binom{n+1}{k}^{-1} \quad \text { for } k=0,1, \ldots, n $$  
Determine $P(n+1)$.
13. Lemma. For any polynomial $P$ of degree at most $n$, 
$$ \sum_{i=0}^{n+1}(-1)^{i}\binom{n+1}{i} P(i)=0 $$
Proof. We shall use induction on $n$. For $n=0$ it is trivial. Assume that it is true for $n=k$ and suppose that $P(x)$ is a polynomial of degree $k+1$. Then $P(x)-P(x+1)$ clearly has degree at most $k$; hence (1) gives 
$$ \begin{aligned} 0 & =\sum_{i=0}^{k+1}(-1)^{i}\binom{k+1}{i}(P(i)-P(i+1)) \\ & =\sum_{i=0}^{k+1}(-1)^{i}\binom{k+1}{i} P(i)+\sum_{i=1}^{k+2}(-1)^{i}\binom{k+1}{i-1} P(i) \\ & =\sum_{i=0}^{k+2}(-1)^{i}\binom{k+2}{i} P(i) \end{aligned} $$
This completes the proof of the lemma. Now we apply the lemma to obtain the value of $P(n+1)$. Since $P(i)=$ $\binom{n+1}{i}^{-1}$ for $i=0,1, \ldots, n$, we have 
$$ 0=\sum_{i=0}^{n+1}(-1)^{i}\binom{n+1}{i} P(i)=(-1)^{n+1} P(n+1)+ \begin{cases}1, & 2 \mid n ; \\ 0, & 2 \nmid n .\end{cases} $$
It follows that $P(n+1)= \begin{cases}1, & 2 \mid n ; \\ 0, & 2 \nmid n .\end{cases}$