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6
a_{n+1} =\frac{a_n^2}{a_n +1} , prove [a_n] = 1994- n
a_{n+1} =\frac{a_n^2}{a_n +1} , prove [a_n] = 1994- n
Source: Slovenia TST 1998 p6
February 15, 2020
Sequence
recurrence relation
algebra
Problem Statement
Let
a
0
=
1998
a_0 = 1998
a
0
=
1998
and
a
n
+
1
=
a
n
2
a
n
+
1
a_{n+1} =\frac{a_n^2}{a_n +1}
a
n
+
1
=
a
n
+
1
a
n
2
for each nonnegative integer
n
n
n
. Prove that
[
a
n
]
=
1994
−
n
[a_n] = 1994- n
[
a
n
]
=
1994
−
n
for
0
≤
n
≤
1000
0 \le n \le 1000
0
≤
n
≤
1000
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