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4
a_n = a_{n-1} - 1/n+ 1/(n + 1) - 1/(n + 2) (2019 Auckland MO J4)
a_n = a_{n-1} - 1/n+ 1/(n + 1) - 1/(n + 2) (2019 Auckland MO J4)
Source:
November 14, 2021
algebra
Sequence
Problem Statement
Suppose
a
1
=
1
6
a_1 =\frac16
a
1
=
6
1
and
a
n
=
a
n
−
1
−
1
n
+
2
n
+
1
−
1
n
+
2
a_n = a_{n-1} - \frac{1}{n}+ \frac{2}{n + 1} - \frac{1}{n + 2}
a
n
=
a
n
−
1
−
n
1
+
n
+
1
2
−
n
+
2
1
for
n
>
1
n > 1
n
>
1
. Find
a
100
a_{100}
a
100
.
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