25th BMO 1989 Problem 3 f(x) is a polynomial of degree n such that f(0) = 0, f(1) = 1/2, f(2) = 2/3, f(3) = 3/4, ... , f(n) = n/(n+1). Find f(n+1).
Put g(x) = (x+1)f(x)-x. Then g(0) = g(1) = ... = g(n) = 0, so g(x) = Ax(x-1)(x-2)...(x-n). g(-1) = 1, so A = (-1) n+1 /n+1!. Hence g(n+1) = (-1) n+1 . So if n is odd, g(n+1) = 1 and f(n+1) = 1. If n is even, then g(n+1) = -1 and f(n+1) = n/(n+2). Thanks to Suat Namli 25th BMO 1989 © John Scholes jscholes@kalva.demon.co.uk 8 February 2004 Last corrected/updated 8 Feb 04