[Thụy An]

Nguyên văn bởi hung.vx

Chứng minh rằng
$$\sum\limits_{k=1}^{n} \binom {2n-k} {n} = \binom {2n} {n-1}.$$

Kosovo Mathematical Olympiad 2018

Xuất phát từ hai đồng nhất thức quen thuộc
\[\left( \begin{array}{c}
m\\
k
\end{array} \right) = \left( \begin{array}{c}
m - 1\\
k
\end{array} \right) + \left( \begin{array}{c}
m - 1\\
k - 1
\end{array} \right) =\binom {m} {m-k} .\]
Ta có
\[\begin{array}{l}
\left( \begin{array}{c}
2n\\
n - 1
\end{array} \right) &= \left( \begin{array}{c}
2n - 1\\
n - 1
\end{array} \right) + \left( \begin{array}{c}
2n - 1\\
n - 2
\end{array} \right)\\
&= \left( \begin{array}{c}
2n - 1\\
n - 1
\end{array} \right) + \left( \begin{array}{c}
2n - 2\\
n - 2
\end{array} \right) + \left( \begin{array}{c}
2n - 2\\
n - 3
\end{array} \right)\\
& = \left( \begin{array}{c}
2n - 1\\
n - 1
\end{array} \right) + \left( \begin{array}{c}
2n - 2\\
n - 2
\end{array} \right) + \left( \begin{array}{c}
2n - 3\\
n - 3
\end{array} \right) + \left( \begin{array}{c}
2n - 3\\
n - 4
\end{array} \right)\\
&= ...\\
& = \left( \begin{array}{c}
2n - 1\\
n - 1
\end{array} \right) + \left( \begin{array}{c}
2n - 2\\
n - 2
\end{array} \right) + ... + \left( \begin{array}{c}
n\\
0
\end{array} \right)\\
&= \left( \begin{array}{c}
2n - 1\\
n
\end{array} \right) + \left( \begin{array}{c}
2n - 2\\
n
\end{array} \right) + ... + \left( \begin{array}{c}
n\\
n
\end{array} \right).
\end{array}\]
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