《计算机中的数学》（第五章）习题

课后习题（部分） （个人答案，非标准答案！）

习题5.3：

解：

Proof：数学归纳法：
         ~~~~~~~~         令$P\left(n\right)$为：

$$\begin{array}{r}{1}^3+2^3+\cdots +n^3={\left(\frac{n\left(n+1\right)}{2}\right)}^2\left(\forall n\ge 1\right)\end{array}$$
归纳基础：$\because$ 当$n=1$时，
$$\begin{array}{r}{1}^3={\left(\frac{1\left(1+1\right)}{2}\right)}^2=1\end{array}$$
$\therefore P\left(1\right)$ 为真；

归纳假设：设$P\left(k\right)$为真，即：
$$\begin{array}{r}{1}^3+2^3+\cdots +k^3={\left(\frac{k\left(k+1\right)}{2}\right)}^2\left(k\in n\right)\end{array}$$
归纳步骤：当$n=k+1$时，方程两边同时加上${\left(k+1\right)}^3$，
$$\begin{array}{rl}{1}^3+2^3+\cdots +k^3+{\left(k+1\right)}^3& ={\left(\frac{k\left(k+1\right)}{2}\right)}^2+{\left(k+1\right)}^3\\ & =\frac{k^2{\left(k+1\right)}^2}{4}+\frac{\left(4k+4\right){\left(k+1\right)}^2}{4}\\ & ={\left(\frac{\left(k+1\right)\left(k+2\right)}{2}\right)}^2\end{array}$$
         ~~~~~~~~         因此，可知$P\left(n\right)$蕴涵$P\left(n+1\right)$，
综上所述，对于任意$n\ge 1$，$P\left(n\right)$都为真。

习题5.4：

解：

Proof：数学归纳法：
         ~~~~~~~~         令$P\left(n\right)$为：

$$\begin{array}{r}1+r+r^2+\cdots +r^n=\frac{r^{n+1}-1}{r-1}\left(\forall n\in N,r\ne 1\right)\end{array}$$
归纳基础：$\because$ 当$n=1$时，
$$\begin{array}{r}1+r^1=\frac{r^2-1}{r-1}=r+1\end{array}$$
$\therefore P\left(1\right)$ 为真；

归纳假设：设$P\left(k\right)$为真，即：
$$\begin{array}{r}1+r+r^2+\cdots +r^k=\frac{r^{k+1}-1}{r-1}\left(k\in N,r\ne 1\right)\end{array}$$
归纳步骤：当$n=k+1$时，方程两边同时加上$r^{k+1}$，
$$\begin{array}{rl}1+r+r^2+\cdots +r^k+r^{k+1}& =\frac{r^{k+1}-1}{r-1}+r^{k+1}\\ & =\frac{r^{k+1}-1}{r-1}+\frac{r^{k+1}\times r-r^{k+1}}{r-1}\\ & =\frac{r^{k+2}-1}{r-1}\end{array}$$
         ~~~~~~~~         因此，可知$P\left(n\right)$蕴涵$P\left(n+1\right)$，
综上所述，对于任意$n\in N$, 且$r\ne 1$，$P\left(n\right)$都为真。

习题5.5：

解：

Proof：数学归纳法：
         ~~~~~~~~         令$P\left(n\right)$为：

$$\begin{array}{r}1+\frac{1}{4}+\frac{1}{9}+\cdots +\frac{1}{n^2}=2-\frac{1}{n}\left(\forall n>1\right)\end{array}$$
归纳基础：$\because$ 当$n=2$时，
$$\begin{array}{r}1+\frac{1}{4}=\frac{5}{4}<2-\frac{1}{2}=\frac{6}{4}\end{array}$$
$\therefore P\left(1\right)$ 为真；

归纳假设：设$P\left(k\right)$为真，即：
$$\begin{array}{r}1+\frac{1}{4}+\frac{1}{9}+\cdots +\frac{1}{k^2}=2-\frac{1}{k}\left(k\in n\right)\end{array}$$
归纳步骤：当$n=k+1$时，方程两边同时加上${\left(k+1\right)}^3$，
$$\begin{array}{rl}1+\frac{1}{4}+\frac{1}{9}+\cdots +\frac{1}{k^2}+\frac{1}{{\left(k+1\right)}^2}& <2-\frac{1}{k}+\frac{1}{{\left(k+1\right)}^2}\left(k\in n\right)\\ & <2-\frac{{\left(k+1\right)}^2-k}{k\left(k+1\right)}\frac{1}{k+1}\\ & <2-\frac{k^2+k+1}{k^2+k}\frac{1}{k+1}\left(\frac{k^2+k+1}{k^2+k}>1\right)\\ & <2-\frac{1}{k+1}\end{array}$$
         ~~~~~~~~         因此，可知$P\left(n\right)$蕴涵$P\left(n+1\right)$，
综上所述，对于任意$ n>1$, $P\left(n\right)$都为真。

习题5.8：

解：

Proof：强数学归纳法：
         ~~~~~~~~         令$P\left(n\right)$为：

$$\begin{array}{r}F\left(n-1\right)\cdot F\left(n+1\right)-F{\left(n\right)}^2={\left(-1\right)}^n\left(\forall n\ge 1\right)\end{array}$$
归纳基础：$\because$ 由斐波那契数列可知：$F\left(0\right)=0,F\left(1\right)=1$
                     ~~~~~~~~~~~~~~~~~~~~                     当$n=1$时，$F\left(2\right)=1$，
$$\begin{array}{r}0\times 1-1^2={\left(-1\right)}^1=-1\end{array}$$
                     ~~~~~~~~~~~~~~~~~~~~                     当$n=2$时，$F\left(3\right)=2$，
$$\begin{array}{r}1\times 2-1^2={\left(-1\right)}^2=1\end{array}$$
                     ~~~~~~~~~~~~~~~~~~~~                     当$n=3$时，$F\left(4\right)=3$，
$$\begin{array}{r}1\times 3-2^2={\left(-1\right)}^3=-1\end{array}$$
$\therefore P\left(1\right),P\left(2\right),P\left(3\right)$ 为真，且${\left(-1\right)}^n=-{\left(-1\right)}^{n+1},{\left(-1\right)}^n={\left(-1\right)}^{n+2}$；

归纳假设：设$P\left(k\right)$为真，且在集合$C=\left\{1,2,3,\cdots ,k\right\}$中$\forall m\in C$,都有$P\left(m\right)$为真，即：
$$\begin{array}{r}F\left(m-1\right)\cdot F\left(m+1\right)-F{\left(m\right)}^2={\left(-1\right)}^m\left(\forall m\in C\right)\end{array}$$
归纳步骤：当$n=k+1$时，
$$\begin{array}{rl}& F\left(k\right)\cdot F\left(k+2\right)-F{\left(k+1\right)}^2\\ =& \left[F\left(k-2\right)+F\left(k-1\right)\right]\left[F\left(k+1\right)+F\left(k\right)\right]-{\left[F\left(k\right)+F\left(k-1\right)\right]}^2\\ =& F\left(k-2\right)\cdot F\left(k+1\right)+F\left(k-2\right)\cdot F\left(k\right)+F\left(k-1\right)\cdot F\left(k+1\right)+F\left(k-1\right)\cdot F\left(k\right)\\ & -F{\left(k\right)}^2-2F\left(k\right)\cdot F\left(k-1\right)-F{\left(k-1\right)}^2\\ =& \left[F\left(k-1\right)\cdot F\left(k+1\right)-F{\left(k\right)}^2\right]+\left[F\left(k-2\right)\cdot F\left(k\right)-F{\left(k-1\right)}^2\right]\\ & +F\left(k-2\right)\cdot F\left(k+1\right)-F\left(k\right)\cdot F\left(k-1\right)\\ =& \left[F\left(k-1\right)\cdot F\left(k+1\right)-F{\left(k\right)}^2\right]+\left[F\left(k-2\right)\cdot F\left(k\right)-F{\left(k-1\right)}^2\right]\\ & +F\left(k-2\right)\cdot \left[F\left(k\right)+F\left(k-1\right)\right]-F\left(k-1\right)\cdot \left[F\left(k-2\right)+F\left(k-1\right)\right]\\ =& \left[F\left(k-1\right)\cdot F\left(k+1\right)-F{\left(k\right)}^2\right]+\left[F\left(k-2\right)\cdot F\left(k\right)-F{\left(k-1\right)}^2\right]\\ & +\left[F\left(k-2\right)\cdot F\left(k\right)-F{\left(k-1\right)}^2\right]\\ =& {\left(-1\right)}^k+{\left(-1\right)}^{k-1}+{\left(-1\right)}^{k-1}\\ =& {\left(-1\right)}^{k-1}\\ =& {\left(-1\right)}^{k+1}\end{array}$$
         ~~~~~~~~         可见，可知$P\left(1\right),\cdots ,P\left(n\right)$蕴涵$P\left(n+1\right)$，证毕。
综上所述，由强归纳法可知，原命题为真。

习题5.11：

解：

Proof：强数学归纳法：
         ~~~~~~~~         令$P\left(n\right)$为：不管有多少个小方格、不论怎么摆放，得到的周长一定是偶数。

         ~~~~~~~~         其中$n$为小方格数量，1个小方格周长$c_0$为4。
归纳基础：$\because$ 当$n=1$时，放入第$1$个小方格没有重叠的边，
$$\begin{array}{r}{c}_1=c_0=4\end{array}$$
                     ~~~~~~~~~~~~~~~~~~~~                     当$n=2$时，放入第$2$个小方格有一条重叠的边，

$$\begin{array}{r}{c}_2=c_0-1+3=6\end{array}$$
                     ~~~~~~~~~~~~~~~~~~~~                     当$n=4$时，放入第$4$个小方格有两条重叠的边，
$$\begin{array}{r}{c}_4=c_0+\left(-1+3\right)\times 2-2+2=8\end{array}$$
                     ~~~~~~~~~~~~~~~~~~~~                     当$n=6$时，放入第$6$个小方格有三条重叠的边，
$$\begin{array}{r}{c}_6=c_0+\left(-1+3\right)\times 4-3+1=10\end{array}$$
                     ~~~~~~~~~~~~~~~~~~~~                     当$n=9$时，放入第$9$个小方格有四条重叠的边，

$$\begin{array}{r}{c}_9=c_0+\left(-1+3\right)\times 7-4=14\end{array}$$
$\therefore P\left(1\right),P\left(2\right),P\left(4\right),P\left(6\right),P\left(9\right)$ 为真；

归纳假设：设$P\left(k\right)$为真，且在集合$T=\left\{1,2,3,\cdots ,k\right\}$中$\forall m\in T$,都有$P\left(m\right)$为真，
                     ~~~~~~~~~~~~~~~~~~~~                     即：加入第m个小方格后周长仍为偶数；

归纳步骤：当$n=k+1$时，加入第$k+1$个新小方格时周长为$c_{k+1}$，
$$\begin{array}{rl}{c}_{k+1}& =c_k-1+3=c_k+2\\ c_{k+1}& =c_k-2+2=c_k\\ c_{k+1}& =c_k-3+1=c_k-2\\ c_{k+1}& =c_k-4\end{array}$$
                     ~~~~~~~~~~~~~~~~~~~~                     其中$\left(1\right)$为重叠一条边，$\left(2\right)$为重叠两条边，$\left(3\right)$为重叠三条边，$\left(4\right)$为重叠四条边，
                     ~~~~~~~~~~~~~~~~~~~~                     且均为偶数；
         ~~~~~~~~         可见，可知$P\left(1\right),\cdots ,P\left(n\right)$蕴涵$P\left(n+1\right)$，证毕。
综上所述，由强归纳法可知，原命题为真。

习题5.12：

解：

Proof：数学归纳法：
         ~~~~~~~~         令$P\left(n\right)$为：

$$\begin{array}{r}A\cap \bigcup _{i=1}^n{B}_i=\bigcup _{i=1}^n\left(A\cap B_i\right)\end{array}$$
归纳基础：$\because$ 当$n=1$时，
$$\begin{array}{rl}A\cap \bigcup _{i=1}^1{B}_i& =\bigcup _{i=1}^1\left(A\cap B_i\right)\\ A\cap B_1& =A\cap B_1\end{array}$$
$\therefore P\left(1\right)$ 为真；

归纳假设：设$P\left(k\right)$为真，即：
$$\begin{array}{r}A\cap \bigcup _{i=1}^k{B}_i=\bigcup _{i=1}^k\left(A\cap B_i\right)\end{array}$$
归纳步骤：当$n=k+1$时，设$z\in \left(A\cap {\bigcup }_{i=1}^k{B}_i\right)$，即：
$$\begin{array}{rl}& z\in \left[A\cap \left({\bigcup }_{i=1}^k{B}_i\cup B_{k+1}\right)\right]\\ \iff & \left(z\in A\right)\wedge \left[\left(z\in {\bigcup }_{i=1}^k{B}_i\right)\vee \left(z\in B_{k+1}\right)\right]\\ \iff & \left[\left(z\in A\right)\wedge \left(z\in {\bigcup }_{i=1}^k{B}_i\right)\right]\vee \left[\left(z\in A\right)\wedge \left(z\in B_{k+1}\right)\right]\\ \iff & \left[z\in \left(A\cap {\bigcup }_{i=1}^k{B}_i\right)\right]\vee \left[z\in \left(A\cap B_{k+1}\right)\right]\\ \iff & z\in \left[\left(A\cap {\bigcup }_{i=1}^k{B}_i\right)\cup \left(A\cap B_{k+1}\right)\right]\\ \iff & z\in \left[\left(A\cap B_1\right)\cup \left(A\cap B_1\right)\cup \cdots \cup \left(A\cap B_k\right)\cup \left(A\cap B_{k+1}\right)\right]\\ \iff & z\in {\bigcup }_{i=1}^{k+1}\left(A\cap B_i\right)\end{array}$$
         ~~~~~~~~         因此，可知$P\left(n\right)$蕴涵$P\left(n+1\right)$，
综上所述，对于任意$ n\ge1$, $P\left(n\right)$都为真。

习题5.14：

解：

Proof：数学归纳法：
         ~~~~~~~~         令$P\left(n\right)$为：

$$\begin{array}{r}\sum _1^{n}k\cdot k!=\left(n+1\right)!-1\left(\forall n\ge 1\right)\end{array}$$
归纳基础：$\because$ 当$n=1$时，
$$\begin{array}{r}\sum _1^{1}k\cdot k!=1\times 1=\left(1+1\right)!-1=1\end{array}$$
$\therefore P\left(1\right)$ 为真；

归纳假设：设$P\left(m\right)$为真，即：
$$\begin{array}{r}\sum _1^{m}k\cdot k!=\left(m+1\right)!-1\left(m\ge 1\right)\end{array}$$
归纳步骤：当$n=m+1$时，方程两边同时加上$\left(m+1\right)\left(m+1\right)!$，
$$\begin{array}{rl}\sum _1^{m}k\cdot k!+\left(m+1\right)\left(m+1\right)!& =\left(m+1\right)!-1+\left(m+1\right)\left(m+1\right)!\\ \sum _1^{m+1}k\cdot k!& =\left(m+2\right)!-1\end{array}$$
         ~~~~~~~~         因此，可知$P\left(n\right)$蕴涵$P\left(n+1\right)$，
综上所述，对于任意$ n\ge1$, $P\left(n\right)$都为真。

习题5.15：

解：

Proof：数学归纳法：
         ~~~~~~~~         令$P\left(n\right)$为：

$$\begin{array}{r}{0}^3+1^3+2^3+\cdots +n^3={\left(\frac{n\left(n+1\right)}{2}\right)}^2\left(\forall n\ge 0\right)\end{array}$$
归纳基础：$\because$ 当$n=0$时，
$$\begin{array}{r}{0}^3={\left(\frac{0\left(0+1\right)}{2}\right)}^2=0\end{array}$$
$\therefore P\left(0\right)$ 为真；

归纳假设：设$P\left(k\right)$为真，即：
$$\begin{array}{r}{0}^3+1^3+2^3+\cdots +k^3={\left(\frac{k\left(k+1\right)}{2}\right)}^2\left(k\in n\right)\end{array}$$
归纳步骤：当$n=k+1$时，方程两边同时加上${\left(k+1\right)}^3$，
$$\begin{array}{rl}{0}^3+1^3+2^3+\cdots +k^3+{\left(k+1\right)}^3& ={\left(\frac{k\left(k+1\right)}{2}\right)}^2+{\left(k+1\right)}^3\\ & =\frac{k^2{\left(k+1\right)}^2}{4}+\frac{\left(4k+4\right){\left(k+1\right)}^2}{4}\\ & ={\left(\frac{\left(k+1\right)\left(k+2\right)}{2}\right)}^2\end{array}$$
         ~~~~~~~~         因此，可知$P\left(n\right)$蕴涵$P\left(n+1\right)$，
综上所述，对于任意$n\ge 0$，$P\left(n\right)$都为真。

习题5.19：

解：

Proof：数学归纳法：
         ~~~~~~~~         令$P\left(n\right)$为：

$$\begin{array}{r}1\cdot 2+2\cdot 3+3\cdot 4+\cdots +n\left(n+1\right)=\frac{n\left(n+1\right)\left(n+2\right)}{3}\left(\forall n\ge 1\right)\end{array}$$
归纳基础：$\because$ 当$n=1$时，
$$\begin{array}{r}1\cdot 2=\frac{1\left(1+1\right)\left(1+2\right)}{3}=2\end{array}$$
$\therefore P\left(1\right)$ 为真；

归纳假设：设$P\left(k\right)$为真，即：
$$\begin{array}{r}1\cdot 2+2\cdot 3+3\cdot 4+\cdots +k\left(k+1\right)=\frac{k\left(k+1\right)\left(k+2\right)}{3}\left(k\in n\right))\end{array}$$
归纳步骤：当$n=k+1$时，方程两边同时加上${\left(k+1\right)}^3$，
$$\begin{array}{rl}1\cdot 2+2\cdot 3+3\cdot 4+\cdots +k\left(k+1\right)+\left(k+1\right)\left(k+2\right)& =\frac{k\left(k+1\right)\left(k+2\right)}{3}+\left(k+1\right)\left(k+2\right)\\ & =\frac{k\left(k+1\right)\left(k+2\right)}{3}+\frac{3\left(k+1\right)\left(k+2\right)}{3}\\ & =\frac{\left(k+1\right)\left(k+2\right)\left(k+3\right)}{3}\end{array}$$
         ~~~~~~~~         因此，可知$P\left(n\right)$蕴涵$P\left(n+1\right)$，
综上所述，对于任意$n\ge 1$，$P\left(n\right)$都为真。

----以上为个人思考与见解，有误请指点，有想法也可联系交流

               ~~~~~~~~~~~~~~               谢谢观看！