polynomial p(x) with real coefficients

The Problem:
	The polynomial p(x) with real coefficients has degree 2011 and satisfies p(n)=nn+1 for all integers n,0≤n≤2011 . Compute p(2012) .

Solution: p(2012)=1.

Define q(x):=(x+1)p(x)−x . Note that q(x) is a polynomial of degree 2012, and we know all 2012 of its zeros, namely q(0)=q(1)=q(2)…=q(2011)=0 . We can therefore write it as a product of its linear factors,

q(x)=Cx(x−1)...(x−2011),

for some constant C . To find C note that by its definition,

q(−1)=(−1+1)p(−1)−(−1)=1,

while in factored form,

q(−1)=C(−1)(−2)…(−2012)=C⋅2012!;

therefore,

C=12012!.

Because
p(x)=1x+1(q(x)+x) , we find that

p(2012)=12013(12012!⋅(2012⋅2011⋅…⋅1)+2012)=20132013=1.