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\(C_n^{n - 1} \cdot C_{n + 1}^n\) 的值;
参考答案:\({n^2} + n\)
\(C_n^{n - 1} \cdot C_{n + 1}^n \)
\(= C_n^{n - 1} \cdot \left( {C_n^n + C_n^{n - 1}} \right) \)
\(= C_n^1 \cdot \left( {1 + C_n^1} \right) = {n^2} + n\)(或原式\( = C_n^1 \cdot C_{n + 1}^1 \)
\(= n\left( {n + 1} \right) = {n^2} + n\)).