Question 31
Consider $1+2+\cdots+n=\tfrac12n(n+1)$. If $P_1(x)=\tfrac12x(x+1)$, then it is the unique polynomial such that $P_1(n)=1+2+\cdots+n$ for all positive integers $n$. In general, for every positive integer $k$, there is a unique polynomial $P_k(x)$ such that
$$P_k(n)=1^k+2^k+3^k+\cdots+n^k\quad(n=1,2,\ldots).$$
Find $P_{2010}(-\tfrac12)$.
$$P_k(n)=1^k+2^k+3^k+\cdots+n^k\quad(n=1,2,\ldots).$$
Find $P_{2010}(-\tfrac12)$.