8th CanMO 1976 Problem 2 The real sequence x 0 , x 1 , x 2 , ... is defined by x 0 = 1, x 1 = 2, n(n+1) x n+1 = n(n-1) x n - (n-2) x n-1 . Find x 0 /x 1 + x 1 x 2 + ... + x 50 /x 51 .
We show by induction that x n = 1/n! for n ≥ 2. We have 1·2 x 2 = 1·0. x 1 - (-1) x 0 = 1, so x 2 = 1/2 and the result is true for n = 2. We also have 2·3 x 3 = 2·1 x 2 - 0. x 1 = 1, so x 3 = 1/6 and the result is true for n = 3. Suppose it is true for n and n-1. Then n(n+1) x n+1 = n(n-1)/n! - (n-2)/(n-1)! = 1/(n-2)! - 1/( (n-1) (n-3)! ) = (n-1 - (n-2) )/(n-1)! = 1/(n-1)!, so x n+1 = 1/(n+1)! and the result is true for n+1. Thus x 0 /x 1 + x 1 x 2 + ... + x 50 /x 51 = 1/2 + 2/(1/2) + (1/2)/(1/6) + ... + (1/50!)/(1/51!) = 1/2 + 4 + 3 + 4 + 5 + ... + 51 = 3/2 + (1 + 2 + ... + 51) = 3/2 + 51.52/2 = 1327 1/2. 8th CanMO 1976 © John Scholes jscholes@kalva.demon.co.uk 10 June 2002