高中数学选择性必修 第二册（381题）

---

知识点：第四章 数列

参考答案：证明(数学归纳法)：对任意的 \(m \in {{\mathbf{N}}^*}\) ，①当 \(n = m\) 时，左边 \( = \left( {m + 1} \right){\rm{C}}_m^m = m + 1\) ，右边 \( = \left( {m + 1} \right){\rm{C}}_{m + 2}^{m + 2} = m + 1\) ，等式成立.②假设 \(n = k\left( {k⩾m} \right)\) 时命题成立，即 \(\left( {m + 1} \right){\rm{C}}_m^m + \left( {m + 2} \right){\rm{C}}_{m + 1}^m + \left( {m + 3} \right){\rm{C}}_{m + 2}^m + \cdots + \) \(k{\rm{C}}_{k - 1}^m + \left( {k + 1} \right){\rm{C}}_k^m = \left( {m + 1} \right){\rm{C}}_{k + 2}^{m + 2}\) ，当 \(n = k + 1\) 时，左边\( = \left( {m + 1} \right){\rm{C}}_m^m + \left( {m + 2} \right){\rm{C}}_{m + 1}^m + \left( {m + 3} \right){\rm{C}}_{m + 2}^m + \cdots + \)\(k{\rm{C}}_{k - 1}^m + \left( {k + 1} \right){\rm{C}}_k^m + \left( {k + 2} \right){\rm{C}}_{k + 1}^m\) \( = \left( {m + 1} \right){\rm{C}}_{k + 2}^{m + 2} + \left( {k + 2} \right){\rm{C}}_{k + 1}^m\) .又由于右边 \( = \left( {m + 1} \right){\rm{C}}_{k + 3}^{m + 2}\) ，而 \(\left( {m + 1} \right){\rm{C}}_{k + 3}^{m + 2} - \left( {m + 1} \right){\rm{C}}_{k + 2}^{m + 2}{\rm{ = }}\)\(\left( {m + 1} \right)\left[ {\frac{{\left( {k + 3} \right)!}}{{\left( {m + 2} \right)!\left( {k - m + 1} \right)!}} - \frac{{\left( {k + 2} \right)!}}{{\left( {m + 2} \right)!\left( {k - m} \right)!}}} \right]{\rm{ = }}\) \(\left( {m + 1} \right) \times \frac{{\left( {k + 2} \right)!}}{{\left( {m + 2} \right)!\left( {k - m + 1} \right)!}}\left[ {k + 3 - \left( {k - m + 1} \right)} \right]\)\( = \left( {k + 2} \right)\frac{{\left( {k + 1} \right)!}}{{m!\left( {k - m + 1} \right)!}}\)\( = \left( {k + 2} \right){\rm{C}}_{k + 1}^m\) .因此 \(\left( {m + 1} \right){\rm{C}}_{k + 2}^{m + 2} + \left( {k + 2} \right){\rm{C}}_{k + 1}^m = \left( {m + 1} \right){\rm{C}}_{k + 3}^{m + 2}\) ，因此左边\( = \)右边，因此\(n = k + 1\)时命题也成立.综合①②可得命题对任意 \(n⩾m\) 均成立.