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Suppose $P(x)$ is a polynomial of degree $2012$ and $P(x) = 1/x$ when $x$ takes the integer values $1\cdots2013$ (inclusive). What is the value of $P(2014)$?

I get $1/1007$ but I'm not sure if it's right, and my method is very inelegant.

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Note that $x\cdot P(x)$ is a polynomial of degree $2013$ that attains the value $1$ in the $2013$ points $1,2,\dotsc,2013$. So

$$x\cdot P(x) = 1 + c\cdot\prod_{k=1}^{2013} (x-k).$$

Now, $x\cdot P(x)$ evidently attains the value $0$ in $x = 0$, so

$$0 = 1 + c\cdot (-1)^{2013}\cdot 2013!$$

or $c = \dfrac{1}{2013!}$. Then

$$2014\cdot P(2014) = 1 + \frac{1}{2013!}\cdot 2013! = 2,$$

and indeed

$$P(2014) = \frac{2}{2014} = \frac{1}{1007}.$$

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  • 2
    $\begingroup$ What an extremely clever answer! $\endgroup$ – String Jan 16 '14 at 11:49

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