14th Mexico 2000 Problem A2 A triangle is constructed like that below, but with 1, 2, 3, ... , 2000 as the first row. Each number is the sum of the two numbers immediately above. Find the number at the bottom of the triangle. 1 2 3 4 5 3 5 7 9 8 12 16 20 28 48
2 1998 2001 Solution We claim that row n is 2 n-2 (n+1), 2 n-2 (n+1)+2 n-1 , 2 n-2 (n+1)+2·2 n-1 , 2 n-2 (n+1)+3·2 n-1 , ... . It is obvious for n = 2. Suppose it is true for n. Then row n+1 is 2 n-1 (n+2), 2 n-1 (n+2)+2 n , 2 n-1 (n+2)+2·2 n , 2 n- 1 (n+2)+3·2 n , ... and so it is true for n+1 and hence for all n. In particular, row 2000 has first term (and only term if the first row has only 2000 terms) 2 1998 2001. Thanks to Suat Namli 14th Mexican 2000 © John Scholes jscholes@kalva.demon.co.uk 29 February 2004 Last corrected/updated 8 Mar 04