• question_answer Let $p(x)$ be a polynomial of degree 2009 such that $p(x)=\frac{1}{x+1}$. For $x=0,\,1,\,2,\,3,...,\,2009$.  The value of p(2010) is A)  0                                 B)  1                C)  $\frac{1}{2011}$      D)  can't be determined

We have, $...(x-2008)\,(x-2009)\,x$ which is an identity. Also, for $x=-1$ $-1=k\,(-2)\,(-3)\,(-4)\,(-5)\,....\,(-2009)\,(-2010)\,(-1)$$\Rightarrow$           $k=-\frac{1}{(2010)!}$ For $x=2010,$ $\Rightarrow$            $\,p(2010)=0$