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2015 CHMMC (Fall)
2
2015 CHMMC Tiebreaker 2 - a_{n+1} =1/n a_n + a_{n-1}
2015 CHMMC Tiebreaker 2 - a_{n+1} =1/n a_n + a_{n-1}
Source:
March 1, 2024
algebra
recurrence relation
CHMMC
Problem Statement
Let
a
1
=
1
a_1 = 1
a
1
=
1
,
a
2
=
1
a_2 = 1
a
2
=
1
, and for
n
≥
2
n \ge 2
n
≥
2
, let
a
n
+
1
=
1
n
a
n
+
a
n
−
1
.
a_{n+1} =\frac{1}{n} a_n + a_{n-1}.
a
n
+
1
=
n
1
a
n
+
a
n
−
1
.
What is
a
12
a_{12}
a
12
?
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