IMO 1991 SL 11

Origin: AUS

Problem

Prove that  m=0 (−1)m 1991 −m 1991 −m m 

Solution

To start with, observe that n−m
n−m m  = 1 n 
n−m m  +
n−m−1 m−1  . For n = 1, 2, . . . set Sn = [n/2] m=0(−1)m
n−m m  . Using the identity
m k 

m−1 k  +
m−1 k−1  we obtain the following relation for Sn: Sn+1 =  m (−1)m n −m + 1 m 

 m (−1)m n −m m  +  m (−1)m n −m m −1  = Sn −Sn−1. Since the initial members of the sequence Sn are 1, 1, 0, −1, −1, 0, 1, 1, . . ., we thus find that Sn is periodic with period 6. Now the sum from the problem reduces to 1991 1991  − 1991 1990  + 1989  +\cdot \cdot \cdot− 1991 996  + 995 

1991(S1991 −S1989) = 1991(0 −(−1)) = 1991.