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6
2023 Algebra NT #6 a_k =\frac{ka_{k-1}}{a_{k-1} - (k - 1)}
2023 Algebra NT #6 a_k =\frac{ka_{k-1}}{a_{k-1} - (k - 1)}
Source:
February 28, 2024
algebra
Problem Statement
Suppose
a
1
,
a
2
,
.
.
.
,
a
100
a_1, a_2, ... , a_{100}
a
1
,
a
2
,
...
,
a
100
are positive real numbers such that
a
k
=
k
a
k
−
1
a
k
−
1
−
(
k
−
1
)
a_k =\frac{ka_{k-1}}{a_{k-1} - (k - 1)}
a
k
=
a
k
−
1
−
(
k
−
1
)
k
a
k
−
1
for
k
=
2
,
3
,
.
.
.
,
100
k = 2, 3, ... , 100
k
=
2
,
3
,
...
,
100
. Given that
a
20
=
a
23
a_{20} = a_{23}
a
20
=
a
23
, compute
a
100
a_{100}
a
100
.
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