Problem:
A sequence of integers a1,a2,a3,… is chosen so that an=an−1−an−2 for each n≥3. What is the sum of the first 2001 terms of this sequence if the sum of the first 1492 terms is 1985, and the sum of the first 1985 terms is 1492?
Solution:
Calculating the first eight terms of the sequence, one finds that it cycles in blocks of six terms; i.e., for n=1,2,3,…,an+6=an. More specifically,
an=⎩⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎧a1a2a2−a1−a1−a2a1−a2 if n=1,7,13,…, if n=2,8,14,…, if n=3,9,15,…, if n=4,10,16,…, if n=5,11,17,…, if n=6,12,18,…
Since the sum of any six consecutive terms of the sequence is zero, if we let sn be the sum of the first n terms, then
sn=⎩⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎧a1a1+a22a22a2−a1a2−a10 if n=1,7,13,…, if n=2,8,14,…, if n=3,9,15,…,2001,…,1492,…, if n=4,10,16,…,1985,…, if n=5,11,17,…, if n=6,12,18,…
Therefore,
s1985=a2−a1=1492
and
s1492=2a2−a1=1985,
from which a2=493 and s2001=2a2=986.
The problems on this page are the property of the MAA's American Mathematics Competitions