4.

题目：
通过上指标翻转计算出$\left( {\begin{array}{*{20}{c}}{ - 1}\\k\end{array}} \right)$。
解答：
如果$k \ge 0$，那么
\[
\left( {\begin{array}{*{20}{c}}{ - 1}\\k\end{array}} \right) = {( - 1)^k}\left( {\begin{array}{*{20}{c}}{k - ( - 1) - 1}\\k\end{array}} \right) = {( - 1)^k}\left( {\begin{array}{*{20}{c}}k\\k\end{array}} \right) = {( - 1)^k}
\]
如果$k<0$，那么
\[
\left( {\begin{array}{*{20}{c}}{ - 1}\\k\end{array}} \right) = 0
\]

46.

题目：
求出下列和式的闭形式解，其中$n$是正整数。
\[
\sum\limits_k {\left( {\begin{array}{*{20}{c}}{2k - 1}\\k\end{array}} \right)\left( {\begin{array}{*{20}{c}}{4n - 2k - 1}\\{2n - k}\end{array}} \right)\frac{ { { {( - 1)}^{k - 1}}}}{ {(2k - 1)(4n - 2k - 1)}}}
\]
解答：
由公式$(5.69)$可得
\[
{\mathcal B_{ - 1}}(z) = \sum\limits_k {\left( {\begin{array}{*{20}{c}}{2k - 1}\\k\end{array}} \right)\frac{ { { {( - z)}^k}}}{ {1 - 2k}}} = \sum\limits_k {\left( {\begin{array}{*{20}{c}}{2k - 1}\\k\end{array}} \right)\frac{ { { {( - 1)}^{k - 1}}}}{ {2k - 1}}{z^k}}
\]
\[
{\mathcal{B}_{ - 1}}( - z) = \sum\limits_k {\left( {\begin{array}{*{20}{c}}{2k - 1}\\k\end{array}} \right)\frac{ { {z^k}}}{ {1 - 2k}}} = \sum\limits_k {\left( {\begin{array}{*{20}{c}}{2k - 1}\\k\end{array}} \right)\frac{ { - 1}}{ {2k - 1}}{z^k}}
\]
两式相乘得到${\mathcal{B}_{ - 1}}(z){\mathcal{B}_{ - 1}}( - z)$，其中$z^{2n}$项的系数恰好就是
\[
\begin{array}{l}\sum\limits_k {\left( {\begin{array}{*{20}{c}}{2k - 1}\\k\end{array}} \right)\frac{ { { {( - 1)}^{k - 1}}}}{ {2k - 1}} \cdot \left( {\begin{array}{*{20}{c}}{2(2n - k) - 1}\\{2n - k}\end{array}} \right)\frac{ { - 1}}{ {2(2n - k) - 1}}} \\ = - \sum\limits_k {\left( {\begin{array}{*{20}{c}}{2k - 1}\\k\end{array}} \right)\left( {\begin{array}{*{20}{c}}{4n - 2k - 1}\\{2n - k}\end{array}} \right)\frac{ { { {( - 1)}^{k - 1}}}}{ {(2k - 1)(4n - 2k - 1)}}} \end{array}
\]
所以题目所求的和式的闭形式解就是${\mathcal{B}_{ - 1}}(z){\mathcal{B}_{ - 1}}( - z)$的$z^{2n}$项的系数的相反数。
由公式$(5.69)$还可以得到
\[
{\mathcal{B}_{ - 1}}(z) = \frac{ {1 + \sqrt {1 + 4z} }}{2}
\]
\[
{\mathcal{B}_{ - 1}}( - z) = \frac{ {1 + \sqrt {1 - 4z} }}{2}
\]
所以
\[
(2{\mathcal{B}_{ - 1}}(z) - 1)(2{\mathcal{B}_{ - 1}}( - z) - 1) = \sqrt {1 - 16{z^2}}
\]
展开化简可以得到
\[
{\mathcal{B}_{ - 1}}(z){\mathcal{B}_{ - 1}}( - z) = \frac{1}{4}\sqrt {1 - 16{z^2}} + \frac{1}{2}{\mathcal{B}_{ - 1}}(z) + \frac{1}{2}{\mathcal{B}_{ - 1}}( - z) - 1
\]
而
\[
\begin{array}{l}{(1 - 16{z^2})^{1/2}} = \sum\limits_k {\left( {\begin{array}{*{20}{c}}{1/2}\\k\end{array}} \right){ {( - 16)}^k}{z^{2k}}} \\ = \sum\limits_k {\frac{1}{ {1 - 2k}}\left( {\begin{array}{*{20}{c}}{ - 1/2}\\k\end{array}} \right){ {( - 16)}^k}{z^{2k}}} \\ = \sum\limits_k {\frac{1}{ {1 - 2k}}\frac{ { { {( - 1)}^k}}}{ { {4^k}}}\left( {\begin{array}{*{20}{c}}{2k}\\k\end{array}} \right){ {( - 16)}^k}{z^{2k}}} \\ = \sum\limits_k {\frac{1}{ {1 - 2k}}\left( {\begin{array}{*{20}{c}}{2k}\\k\end{array}} \right){4^k}{z^{2k}}} \end{array}
\]
所以题目答案即${\mathcal{B}_{ - 1}}(z){\mathcal{B}_{ - 1}}( - z)$的$z^{2n}$项的系数的相反数为
\[
\left( {\begin{array}{*{20}{c}}{2n}\\n\end{array}} \right)\frac{ { {4^{n - 1}}}}{ {2n - 1}} + \left( {\begin{array}{*{20}{c}}{4n - 1}\\{2n}\end{array}} \right)\frac{1}{ {4n - 1}}
\]

64.

题目：
计算
\[
\sum\limits_{k = 0}^n {\left( {\begin{array}{*{20}{c}}n\\k\end{array}} \right)/\left\lceil {\frac{ {k + 1}}{2}} \right\rceil }
\]
解答：
\[
\begin{array}{l}\sum\limits_{k = 0}^n {\left( {\begin{array}{*{20}{c}}n\\k\end{array}} \right)/\left\lceil {\frac{ {k + 1}}{2}} \right\rceil } \\ = \sum\limits_{k = 0}^n {\left( {\left( {\begin{array}{*{20}{c}}n\\{2k}\end{array}} \right) + \left( {\begin{array}{*{20}{c}}n\\{2k + 1}\end{array}} \right)} \right)\frac{1}{ {k + 1}}} \\ = \sum\limits_{k = 0}^n {\left( {\begin{array}{*{20}{c}}{n + 1}\\{2k + 1}\end{array}} \right)\frac{1}{ {k + 1}}} \\ = \frac{2}{ {n + 2}}\sum\limits_{k = 0}^n {\left( {\begin{array}{*{20}{c}}{n + 2}\\{2k + 2}\end{array}} \right)} \\ = \frac{ { {2^{n + 2}} - 2}}{ {n + 2}}\end{array}
\]

65.

题目：
证明
\[
\sum\limits_k {\left( {\begin{array}{*{20}{c}}{n - 1}\\k\end{array}} \right){n^{ - k}}(k + 1)!} = n
\]
解答：
等号左边可以写为
\[
\sum\limits_{0 \le k \le n - 1} {\left( {\begin{array}{*{20}{c}}{n - 1}\\k\end{array}} \right){n^{ - k}}(k + 1)!}
\]
替换$k$为$n-1-k$，得到
\[
\sum\limits_{0 \le k \le n - 1} {\left( {\begin{array}{*{20}{c}}{n - 1}\\k\end{array}} \right){n^{1 + k - n}}(n - k)!}
\]
即证
\[
\sum\limits_{0 \le k \le n - 1} {\left( {\begin{array}{*{20}{c}}{n - 1}\\k\end{array}} \right){n^{1 + k - n}}(n - k)!} = n
\]
等式两边同时乘以$n^{n-1}$，即证
\[
\sum\limits_{0 \le k \le n - 1} {\left( {\begin{array}{*{20}{c}}{n - 1}\\k\end{array}} \right){n^k}(n - k)!} = {n^n}
\]
等式左边等于
\[
\begin{array}{l}\sum\limits_{0 \le k \le n - 1} {\left( {\begin{array}{*{20}{c}}{n - 1}\\k\end{array}} \right){n^k}(n - k)!} \\ = (n - 1)!\sum\limits_{0 \le k \le n - 1} {\frac{ { {n^k}(n - k)}}{ {k!}}} \\ = (n - 1)!\sum\limits_{0 \le k \le n - 1} {\left( {\frac{ { {n^{k + 1}}}}{ {k!}} - \frac{ { {n^k}}}{ {(k - 1)!}}} \right)} \\ = (n - 1)!\frac{ { {n^n}}}{ {(n - 1)!}}\\ = {n^n}\end{array}
\]
得证。