Question

Let $P\left(x\right)$ be a polynomial of degree $2010$. Suppose $P\left(n\right)=\frac{n}{1+n}$ for all $n=0,1,2,\dots ,2010.$ Find $P\left(2012\right).$
(correct answer + 5, wrong answer 0)

Answer 1

Define $Q\left(x\right)=(1+x)P\left(x\right)-x$
Then $Q\left(x\right)$ is a polynomial of degree $2011$.
Since $Q\left(0\right)=Q\left(1\right)=Q\left(2\right)=\cdots =Q\left(2010\right)=0,$
$\therefore Q\left(x\right)$ can be written as
$Q\left(x\right)=Ax(x-1)(x-2)\cdots (x-2010)$
$1=Q(-1)=A(-1)(-2)(-3)\cdots (-2011)=-A\times \left(2011\right)!$
$\Rightarrow A=-\frac{1}{\left(2011\right)!}$

Now, $Q\left(2012\right)=A\times \left(2012\right)!$
$=-\frac{1}{\left(2011\right)!}\times \left(2012\right)!$
$=-2012$
And $P\left(2012\right)=\frac{Q\left(2012\right)+2012}{2013}$
$=\frac{-2012+2012}{2013}=0$