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8
2014 Algebra #8
2014 Algebra #8
Source:
July 8, 2022
2014
Algebra Test
Problem Statement
Consider the recurrence relation
a
n
+
3
=
a
n
+
2
a
n
+
1
−
2
a
n
a_{n+3}=\frac{a_{n+2}a_{n+1}-2}{a_n}
a
n
+
3
=
a
n
a
n
+
2
a
n
+
1
−
2
with initial condition
(
a
0
,
a
1
,
a
2
)
=
(
1
,
2
,
5
)
(a_0,a_1,a_2)=(1,2,5)
(
a
0
,
a
1
,
a
2
)
=
(
1
,
2
,
5
)
. Let
b
n
=
a
2
n
b_n=a_{2n}
b
n
=
a
2
n
for nonnegative integral
n
n
n
. It turns out that
b
n
+
2
+
x
b
n
+
1
+
y
b
n
=
0
b_{n+2}+xb_{n+1}+yb_n=0
b
n
+
2
+
x
b
n
+
1
+
y
b
n
=
0
for some pair of real numbers
(
x
,
y
)
(x,y)
(
x
,
y
)
. Compute
(
x
,
y
)
(x,y)
(
x
,
y
)
.
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