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Vojtěch Jarník IMC
2014 VJIMC
Problem 3
a binomial sum, alternating
a binomial sum, alternating
Source: VJIMC 2014 2.3
May 21, 2021
Summation
binomial coefficients
algebra
Problem Statement
Let
k
k
k
be a positive even integer. Show that
∑
n
=
0
k
/
2
(
−
1
)
n
(
k
+
2
n
)
(
2
(
k
−
n
)
+
1
k
+
1
)
=
(
k
+
1
)
(
k
+
2
)
2
.
\sum_{n=0}^{k/2}(-1)^n\binom{k+2}n\binom{2(k-n)+1}{k+1}=\frac{(k+1)(k+2)}2.
n
=
0
∑
k
/2
(
−
1
)
n
(
n
k
+
2
)
(
k
+
1
2
(
k
−
n
)
+
1
)
=
2
(
k
+
1
)
(
k
+
2
)
.
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