

Cho n là số tự nhiên lẻ. Chứng minh rằng:
$\left ( C_{n}^{1} \right )^{2} + 2 * \left ( C_{n}^{2} \right )^{2} + ... + n * \left ( C_{n}^{n} \right )^{2} = \frac{n}{2} * C_{2n}^{n}$
$\left ( C_{n}^{1} \right )^{2} + 2 * \left ( C_{n}^{2} \right )^{2} + ... + n * \left ( C_{n}^{n} \right )^{2} = \frac{n}{2} * C_{2n}^{n}$
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