Maths Olympiad Prep

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Algebra Difficulty 5.2 AIME, harder Prove it United States

Problem:
Let p0(x),p1(x),p2(x),p_{0}(x), p_{1}(x), p_{2}(x), \ldots be polynomials such that p0(x)=xp_{0}(x) = x and for all positive integers nn, ddxpn(x)=pn1(x)\frac{d}{dx} p_{n}(x) = p_{n-1}(x). Define the function p(x):[0,)Rp(x): [0, \infty) \to \mathbb{R} by p(x)=pn(x)p(x) = p_{n}(x) for all x[n,n+1]x \in [n, n+1]. Given that p(x)p(x) is continuous on [0,)[0, \infty), compute
n=0pn(2009) \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 pn(x)=1(n+1)!(x+1)n+11n!xnp_{n}(x) = \frac{1}{(n+1)!}(x+1)^{n+1} - \frac{1}{n!} x^{n}. Thus the sum evaluates to e2010e20091e^{2010} - e^{2009} - 1 by the series expansion of exe^{x}.

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