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Serbia Team Selection Test
1992 Yugoslav Team Selection Test
Problem 2
4-sequence recurrence, periodic
4-sequence recurrence, periodic
Source: Yugoslav TST 1992 P2
May 28, 2021
recurrence relation
Sequences
algebra
Problem Statement
Periodic sequences
(
a
n
)
,
(
b
n
)
,
(
c
n
)
(a_n),(b_n),(c_n)
(
a
n
)
,
(
b
n
)
,
(
c
n
)
and
(
d
n
)
(d_n)
(
d
n
)
satisfy the following conditions:
a
n
+
1
=
a
n
+
b
n
,
b
n
+
1
=
b
n
+
c
n
,
a_{n+1}=a_n+b_n,\enspace\enspace b_{n+1}=b_n+c_n,
a
n
+
1
=
a
n
+
b
n
,
b
n
+
1
=
b
n
+
c
n
,
c
n
+
1
=
c
n
+
d
n
,
d
n
+
1
=
d
n
+
a
n
,
c_{n+1}=c_n+d_n,\enspace\enspace d_{n+1}=d_n+a_n,
c
n
+
1
=
c
n
+
d
n
,
d
n
+
1
=
d
n
+
a
n
,
for
n
=
1
,
2
,
…
n=1,2,\ldots
n
=
1
,
2
,
…
. Prove that
a
2
=
b
2
=
c
2
=
d
2
=
0
a_2=b_2=c_2=d_2=0
a
2
=
b
2
=
c
2
=
d
2
=
0
.
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