18th USAMO 1989 Problem 1 Let a n = 1 + 1/2 + 1/3 + ... + 1/n, b n = a 1 + a 2 + ... + a n , c n = b 1 /2 + b 2 /2 + ... + b n /(n+1). Find b 1988 and c 1988 in terms of a 1989 .
Grouping the 1/r terms, we have b n-1 = (n - 1) + (n/2 - 2/2) + (n/3 - 3/3) + ... + (n/n - n/n) = n a n - n. Hence, in particular, b 1988 = 1989 a 1989 - 1989. Hence also b m /(m+1) = a m+1 - 1, so c n-1 = (a 2 + a 3 + ... + a n ) - (n-1) = b n - n = (n+1) a n+1 - (2n+1). But (n+1) a n+1 = (n+1) a n + 1, so c n-1 = (n+1) a n - 2n. Hence, in particular, c 1988 = 1990 a 1989 - 3978. 18th USAMO 1989 © John Scholes jscholes@kalva.demon.co.uk 11 May 2002