BUMP How has this gone unanswered for two years!!! Should it be moved to a tougher forum? Right now I can't think of anything but its probably tractable.

what? how is this in the least bit difficult, you just need the x coefficient in: $P(x-\frac{21*22}{2})=P_{21}(x)$

What? THat was obvious. I just read it wrong... must be a pretty weak olympiad

so is answer 61744?

me@home wrote: I just read it wrong... Me too. Personally, I read the recursion as $P_{n}(x) = P_{n-1}(x) \cdot (x-n)$, which is a little more interesting, I think, although still not hard. For one thing, quadratic and higher terms can be ignored entirely.

t0rajir0u wrote: me@home wrote: I just read it wrong... Me too. Personally, I read the recursion as $P_{n}(x) = P_{n-1}(x) \cdot (x-n)$, which is a little more interesting, I think, although still not hard. For one thing, quadratic and higher terms can be ignored entirely. if it was $p_{n}(x)=p_{n-1}(x)(x-n)$ $p_{n}(x)=p_{0}(x)\prod_{i=1}^{n}(x-i)=\sum_{k=0}^{n+3}c_{k}x^{k}$ the coefficient of $x$ is $c_{1}=(-1)^{n}(-67+2000\sum_{k=1}^{n}\frac{1}{k})n!$

this seems to be the same problem: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=80691

$P_{21} (x) = (x-(1+2+...+21))^3 +213 (x- (1+2+..+21))^2 -67(x-(1+2+...+21)) +2000$ The coefficient of x is $3(231)^2 -213(231)(2)-67=61610$