Problem:
A sequence of numbers is defined recursively by a1​=1,a2​=73​, and
an​=2an−2​−an−1​an−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)−13​4(n−2)−13​⋅4(n−1)−13​​​=4n−96​−4n−53​4n−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
an​1​=an−2​⋅an−1​2an−2​−an−1​​=an−1​2​−an−2​1​
which is equivalent to
an​1​−an−1​1​=an−1​1​−an−2​1​
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
a2019​1​=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