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2020 BMT Fall
6
BMT Algebra #6 - Sum with Combinations
BMT Algebra #6 - Sum with Combinations
Source:
October 11, 2020
Bmt
algebra
Problem Statement
Given that
(
n
k
)
=
n
!
k
!
(
n
−
k
)
!
\tbinom{n}{k}=\tfrac{n!}{k!(n-k)!}
(
k
n
)
=
k
!
(
n
−
k
)!
n
!
, the value of
∑
n
=
3
10
(
n
2
)
(
n
3
)
(
n
+
1
3
)
\sum_{n=3}^{10}\frac{\binom{n}{2}}{\binom{n}{3}\binom{n+1}{3}}
n
=
3
∑
10
(
3
n
)
(
3
n
+
1
)
(
2
n
)
can be written in the form
m
n
\tfrac{m}{n}
n
m
, where
m
m
m
and
n
n
n
are relatively prime positive integers. Compute
m
+
n
m+n
m
+
n
.
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