31. Consider the identity $1+2+\cdots+n=\frac{1}{2} n(n+1)$. If we set $P_{1}(x)=\frac{1}{2} x(x+1)$, then it is the unique polynomial such that for all positive integer $n, P_{1}(n)=1+2+\cdots+n$. In general, for each positive integer $k$, there is a unique polynomial $P_{k}(x)$ such that
$$
P_{k}(n)=1^{k}+2^{k}+3^{k}+\cdots+n^{k} \quad \text { for each } n=1,2, \ldots .
$$

Find the value of $P_{2010}\left(-\frac{1}{2}\right)$.
31. Ans: 0 .
Let $k$ be a positive even number.

Define $f(x)=P_{k}(x)-P_{k}(x-1)$. Then $f(n)=n^{k}$ for all integer $n \geq 2$. Note that $f$ is a polynomial. We must have $f(x)=x^{k}$. In particular, for integers $n \geq 2$,
$$
\begin{aligned}
P_{k}(-n+1)-P_{k}(-n) & =f(-n+1)=(n-1)^{k}, \\
P_{k}(-n+2)-P_{k}(-n+1) & =f(-n+2)=(n-2)^{k},
\end{aligned}
$$
$$
\begin{aligned}
P_{k}(0)-P_{k}(-1) & =f(0)=0^{k}, \\
P_{k}(1)-P_{k}(0) & =f(1)=1^{k} .
\end{aligned}
$$

Summing these equalities, $P_{k}(1)-P_{k}(-n)=1^{k}+0^{k}+1^{k}+\cdots+(n-1)^{k}$. That is,
$$
P_{k}(-n)+P_{k}(n-1)=0 .
$$

Define $g(x)=P_{k}(-x)+P_{k}(x-1)$. Then $g(n)=0$ for all integers $n \geq 2$. Since $g$ is a polynomial, $g(x)=0$.
In particular, $P_{k}\left(-\frac{1}{2}\right)+P_{k}\left(-\frac{1}{2}\right)=0$, i.e., $P_{k}\left(-\frac{1}{2}\right)=0$.