12th Australian 1991 Problem B2 Let g(n) = (n 2 - 2n + 1) 1/3 + (n 2 - 1) 1/3 + (n 2 + 2n + 1) 1/3 . Find 1/g(1) + 1/g(3) + 1/g(5) + ... + 1/g(999999).
50 Solution Put h = (n+1) 1/3 , k = (n-1) 1/3 , so g(n) = h 2 + hk + k 2 = (h 3 - k 3 )/(h - k) = 2/(h-k) and 1/g(n) = (h-k)/2. So 1/g(1) + 1/g(3) + 1/g(5) + ... + 1/g(999999) = (1/2) (2 1/3 + 4 1/3 - 2 1/3 + 6 1/3 - 4 1/3 + ... + 1000000 1/3 - 999998 1/3 ) = 50. Thanks to Suat Namli Solutions are also available in Australian Mathematical Olympiads 1979-1995 by H Lausch and P J Taylor, ISBN 0858896451, published by Australian Mathematics Trust. It can be ordered from AMT by email. 12th Aus 1991 © John Scholes jscholes@kalva.demon.co.uk 5 February 2004 Last corrected/updated 5 Feb 04