Let $P(x)$ be a polynomial such that for all integers $x \geq 1$,

$$
P(x)=\sum_{n=1}^{x} n^{2012}
$$

(a) Find $P(-2)$.

(b) Find $P(1 / 2)$.
(a) Consider the relation

$$
P(x)=P(x-1)+x^{2012} .
$$

If $x$ is an integer greater than 1, then (1) clearly holds. Therefore (1) holds for all real $x$ (two polynomials cannot have infinitely many common values unless they are the same). We can therefore use the recurrence $P(x-1)=P(x)-x^{2012}$ to compute

$$
\begin{aligned}
P(0) & =P(1)-1^{2012}=1-1=0 \\
P(-1) & =P(0)-0^{2012}=0-0=0 \\
P(-2) & =P(-1)-(-1)^{2012}=0-1=-1 .
\end{aligned}
$$

(b) Consider the relation

$$
P(-1-x)=-P(x) \text {. }
$$

The $x=0$ case of this follows easily from the computations performed in part (a). Given this, we can easily prove (2) by induction for all nonnegative integers $x$ using (1):

$$
\begin{aligned}
P(-2-x) & =P(-1-x)-(-1-x)^{2012} \\
& =-P(x)-(x+1)^{2012} \\
& =-P(x+1)
\end{aligned}
$$

Plugging $x=-1 / 2$ into (2) yields $P(-1 / 2)=0$, so by (1),

$$
P(1 / 2)=P(-1 / 2)+\left(\frac{1}{2}\right)^{2012}=\frac{1}{2^{2012}}
$$