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China National Olympiad
1994 China National Olympiad
5
5
Part of
1994 China National Olympiad
Problems
(1)
China Mathematical Olympiad 1994 problem5
Source: China Mathematical Olympiad 1994 problem5
9/17/2013
For arbitrary natural number
n
n
n
, prove that
∑
k
=
0
n
C
n
k
2
k
C
n
−
k
[
(
n
−
k
)
/
2
]
=
C
2
n
+
1
n
\sum^n_{k=0}C^k_n2^kC^{[(n-k)/2]}_{n-k}=C^n_{2n+1}
∑
k
=
0
n
C
n
k
2
k
C
n
−
k
[(
n
−
k
)
/2
]
=
C
2
n
+
1
n
, where
C
0
0
=
1
C^0_0=1
C
0
0
=
1
and
[
n
−
k
2
]
[\dfrac{n-k}{2}]
[
2
n
−
k
]
denotes the integer part of
n
−
k
2
\dfrac{n-k}{2}
2
n
−
k
.
algebra unsolved
algebra
polynomial
Integer Part