Problem:
Let a sequence be defined as follows: a1=3,a2=3, and for n≥2, an+1an−1=an2+2007. Find the largest integer less than or equal to a2007a2006a20072+a20062.
Solution:
The fact that the equation an+1an−1=an2+2007 holds for n≥2 implies that anan−2=an−12+2007 for n≥3. Subtracting the second equation from the first one yields an+1an−1−anan−2=an2−an−12, or an+1an−1+an−12=anan−2+an2. Dividing the last equation by an−1an and simplifying produces anan+1+an−1= an−1an+an−2. This equation shows that anan+1+an−1 is constant for n≥2. Because a3a1=a22+2007,a3=2016/3=672. Thus anan+1+an−1=3672+3=225, and an+1=225an−an−1 for n≥2. Note that a3=672an>3=a2. Furthermore, if an>an−1, then an+1an−1=an2+2007 implies that
an+1=an−1an2+an−12007=an(an−1an)+an−12007>an+an−12007>an.
Thus by mathematical induction, an>an−1 for all n≥3. Therefore the recurrence an+1=225an−an−1 implies that an+1>225an−an=224an and therefore an≥2007 for n≥4. Finding an+1 from an+1an−1=an2+2007 and substituting into 225=anan+1+an−1 shows that anan−1an2+an−12=225−anan−12007. Thus the largest integer less than or equal to the original fraction is 224.
The problems on this page are the property of the MAA's American Mathematics Competitions