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 inegers. What is p+q ?
Answer Choices:
A. 2020
B. 4039
C. 6057
D. 6061
E. 8078
Solution:
We have an​1​=an−2​⋅an−1​2an−2​−an−1​​=an−1​2​−an−2​1​, in other words, an​1​−an−1​1​=an−1​1​−an−2​1​. So {an​1​} is an arithmetic sequence with step size 37​−1=34​, which means an​1​=1+2018⋅34​=38075​. Since the numerator and the denominator are relatively prime, the answer is (E) 8078 .
OR
It seems reasonable to transform the equation into something else. Let an​=x,an−1​=y, and an−2​=z. Therefore, we have
x=2z−yzy​
2xz−xy=zy
2xz=y(x+z)
y =\dfrac{2 x z}
Thus, y is the harmonic mean of x and z. This implies an​ is a harmonic sequence or equivalently bn​=an​1​ is arithmetic. Now, we have b1​=1,b2​=37​,b3​=311​, and so on. Since the common difference is 34​, we can express bn​ explicitly as bn​=34​(n−1)+1. This gives b2019​=34​(2019−1)+1=38075​ which implies a2019​=80753​=qp​. p+q=(E)8078.
The problems on this page are the property of the MAA's American Mathematics Competitions