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Polynomial recursion, f_k(k)=2^k - OIMU 1998 Problem 5

Source: OIMU 1998 Problem 5

September 12, 2010
algebrapolynomial1998 IberoAmerica Undergrad MOIberoAmerica Undergrad MO

Problem Statement

A sequence of polynomials {fn}n=0∞\{f_n\}_{n=0}^{\infty} is defined recursively by f0(x)=1f_0(x)=1, f1(x)=1+xf_1(x)=1+x, and (k+1)f_{k+1}(x)-(x+1)f_k(x)+(x-k)f_{k-1}(x)=0,   k=1,2,\ldots Prove that fk(k)=2kf_k(k)=2^k for all k≥0k\geq 0.
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