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CHMMC problems
2014 CHMMC (Fall)
4
4
Part of
2014 CHMMC (Fall)
Problems
(2)
2014 Fall Team #4
Source:
3/26/2022
Let
b
1
=
1
b_1 = 1
b
1
=
1
and
b
n
+
1
=
1
+
1
n
(
n
+
1
)
b
1
b
2
.
.
.
b
n
b_{n+1} = 1 + \frac{1}{n(n+1)b_1b_2...b_n}
b
n
+
1
=
1
+
n
(
n
+
1
)
b
1
b
2
...
b
n
1
for
n
≥
1
n \ge 1
n
≥
1
. Find
b
1
2
b_12
b
1
2
.
algebra
2014 CHMMC Tiebreaker 4 - f(i, j, k) = f(i - 1, j + k, 2i - 1)
Source:
3/1/2024
If
f
(
i
,
j
,
k
)
=
f
(
i
−
1
,
j
+
k
,
2
i
−
1
)
f(i, j, k) = f(i - 1, j + k , 2i - 1)
f
(
i
,
j
,
k
)
=
f
(
i
−
1
,
j
+
k
,
2
i
−
1
)
and
f
(
0
,
j
,
k
)
=
j
+
k
f(0, j, k) = j + k
f
(
0
,
j
,
k
)
=
j
+
k
, evaluate
f
(
n
,
0
,
0
)
f(n, 0, 0)
f
(
n
,
0
,
0
)
.
algebra
CHMMC