Problem:
Given that a sequence satisfies x0=0 and ∣xk∣=∣xk−1+3∣ for all integers k≥1, find the minimum possible value of ∣x1+x2+⋯+x2006∣.
Solution:
The condition ∣xk∣=∣xk−1+3∣ is equivalent to xk2=(xk−1+3)2. Thus
k=1∑n+1xk2xn+12k=0∑nxk=k=1∑n+1(xk−1+3)2=k=0∑n(xk+3)2=(k=0∑nxk2)+(6k=0∑nxk)+9(n+1), so =k=1∑n+1xk2−k=0∑nxk2=(6k=0∑nxk)+9(n+1), and =61[xn+12−9(n+1)].
Therefore ∣∣∣∣∑k=12006xk∣∣∣∣=61∣∣∣x20072−18063∣∣∣. Notice that xk is a multiple of 3 for all k, and that xk and k have the same parity. The requested sum will be a minimum when ∣∣∣x20072−18063∣∣∣ is a minimum, that is, when x2007 is the multiple of 3 whose square is as close as possible to 18063. Check odd multiples of 3, and find that 1292<16900,1412>19600, and 1352=18225. The requested minimum is therefore 61∣∣∣1352−18063∣∣∣=27, provided there exists a sequence that satisfies the given conditions and for which x2007=135. An example of such a sequence is
xk=⎩⎪⎪⎨⎪⎪⎧3k−138135 for k≤45 for k>45 and k even for k>45 and k odd
The problems on this page are the property of the MAA's American Mathematics Competitions