Problem:
A sequence of numbers is defined recursively by a1=1,a2=73, and
an=2an−2−an−1an−2⋅an−1
for all n≥3. Then a2019 can be written as qp, where p and q are relatively prime positive integers. What is p+q?
Answer Choices:
A. 2020
B. 4039
C. 6057
D. 6061
E. 8078
Solution:
The sequence begins 1,73,113,153,193,… This pattern leads to the conjecture that an=4n−13. Checking the initial conditions n=1 and n=2, and observing that for n≥3,
2⋅4(n−2)−13−4(n−1)−134(n−2)−13⋅4(n−1)−13=4n−96−4n−534n−93⋅4n−53=6(4n−5)−3(4n−9)9=12n−39=4n−13
confirms the conjecture. Therefore a2019=4⋅2019−13=80753, and the requested sum is 3+8075=(E)8078.
OR
Taking the reciprocal of both sides of the recurrence gives
an1=an−2⋅an−12an−2−an−1=an−12−an−21
which is equivalent to
an1−an−11=an−11−an−21
Thus the sequence of reciprocals is an arithmetic sequence. Its first term is 1, and its common difference is 37−1=34. Its 2019th term is
a20191=1+2018⋅34=38075
so a2019=80753, and the requested sum is 3+8075=(E)8078.
The problems on this page are the property of the MAA's American Mathematics Competitions