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 inegers. What is p+q ?
Answer Choices:
A. 2020
B. 4039
C. 6057
D. 6061
E. 8078
Solution:
We have an1=an−2⋅an−12an−2−an−1=an−12−an−21, in other words, an1−an−11=an−11−an−21. So {an1} is an arithmetic sequence with step size 37−1=34, which means an1=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=an1 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