15. Let $P(x)$ be a polynomial of degree 2010. Suppose $P(n)=\frac{n}{1+n}$ for all $n=0,1,2, \ldots, 2010$. Find $P(2012)$.
15. Answer: 0.
Define $Q(x)=(1+x) P(x)-x$. Then $Q(x)$ is a polynomial of degree 2011 . Since $Q(0)=$ $Q(1)=Q(2)=\cdots=Q(2010)=0$, we can write, for some constant $A$,
$$
Q(x)=A x(x-1)(x-2) \cdots(x-2010) .
$$
$1=Q(-1)=A(-1)(-2)(-3) \cdots(-2011)=-A \cdot 2011!$. Then $Q(2012)=A \cdot 2012!=-2012$, and $P(2012)=\frac{Q(2012)+2012}{2013}=0$.