63rd Putnam 2002 Problem A1 k and n are positive integers. Let f(x) = 1/(x k - 1). Let p(x) = (x k - 1) n+1 f n (x), where f n is the nth derivative. Find p(1). Solution
: (-1) n+1 n! k n . A routine differentiation. Let p n (x) = (x k - 1) n+1 f n (x). So f n = p n /(x k - 1) n+1 . Differentiating, f n+1 = (p n ' (x k - 1) - (n+1)kx k-1 p n )/(x k - 1) n+2 . Hence p n+1 (1) = - (n+1)kp n (1). Also p 1 (1) = -k. Hence p n (1) = (-1) n n! k n . 63rd Putnam 2002 © John Scholes jscholes@kalva.demon.co.uk 11 Dec 2002 Last corrected/updated 16 Dec 02