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1980 Polish MO Finals
6
\sum_{s=n}^{2n} 2^{2n-s}{s \choose n}= 2^{2n}
\sum_{s=n}^{2n} 2^{2n-s}{s \choose n}= 2^{2n}
Source: Polish MO Finals 1980 p6
August 24, 2024
algebra
binomial coefficients
Problem Statement
Prove that for every natural number
n
n
n
we have
∑
s
=
n
2
n
2
2
n
−
s
(
s
n
)
=
2
2
n
.
\sum_{s=n}^{2n} 2^{2n-s}{s \choose n}= 2^{2n}.
s
=
n
∑
2
n
2
2
n
−
s
(
n
s
)
=
2
2
n
.
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