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P_{2008}(2008) =? P_n(x)=P_{n-1}(x)+P_{n-1}(x-1), P_1(x)=2/x

Source: Mathcenter Contest / Oly - Thai Forum 2008 R2 p5 https://artofproblemsolving.com/community/c3196914_mathcenter_contest

November 11, 2022
algebrapolynomialrecurrence relation

Problem Statement

Let P1(x)=1xP_1(x)=\frac{1}{x} and Pn(x)=Pn−1(x)+Pn−1(x−1)P_n(x)=P_{n-1}(x)+P_{n-1}(x-1) for every natural n n greater than 11. Find the value of P2008(2008)P_{2008}(2008).
(Mathophile)
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