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Harvard-MIT Mathematics Tournament

Algebra Difficulty 5.2 AIME, harder Prove it United States

Problem:
Let $p_{0}(x), p_{1}(x), p_{2}(x), \ldots$ be polynomials such that $p_{0}(x) = x$ and for all positive integers $n$ , $\frac{d}{dx} p_{n}(x) = p_{n-1}(x)$ . Define the function $p(x): [0, \infty) \to \mathbb{R}$ by $p(x) = p_{n}(x)$ for all $x \in [n, n+1]$ . Given that $p(x)$ is continuous on $[0, \infty)$ , compute
$\sum_{n=0}^{\infty} p_{n}(2009)$

Solution

Solution:
By writing out the first few polynomials, one can guess and then show by induction that $p_{n}(x) = \frac{1}{(n+1)!}(x+1)^{n+1} - \frac{1}{n!} x^{n}$ . Thus the sum evaluates to $e^{2010} - e^{2009} - 1$ by the series expansion of $e^{x}$ .

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