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

Correct Answer: A

Solution :

 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 \[x=2010,\] \[\Rightarrow \]            \[\,p(2010)=0\]