具体数学习题解答——第2章 和式 Part1

书籍介绍：

https://book.douban.com/subject/21323941/

github仓库：

https://github.com/Doraemonzzz/Concrete-Mathematics

这次回顾第二章，和式。

第2章 和式

热身题

1

$$\sum_{k=4}^{0} q_{k} =0$$

2

$$\begin{aligned} x \times([x>0]-[x<0]) &=\begin{cases} x & x > 0\\ 0 & x= 0\\ -x & x < 0 \end{cases}\\ \end{aligned}$$

因此

$$x \times([x>0]-[x<0]) = |x|$$

3

$$\begin{aligned} \sum_{0 \leq k \leq 5} a_{k}&= a_0 + a_1+a_2+a_3+a_4+a_5\\ \sum_{0 \leq k^2 \leq 5} a_{k^2}&= a_{4} + a_{1}+a_0 + a_1+a_4 \end{aligned}$$

注意第二个式子中满足条件的$k$为$0,\pm 1,\pm2$。

4

a

$$\begin{aligned} \sum_{1 \leq i<j<k \leq 4} a_{i j k} &= \sum_{i=1}^4 \sum_{j=i+1}^{4} \sum_{k=j+1}^4 a_{ijk}\\ &= ((a_{123} + a_{124}) + a_{134}) + a_{234} \end{aligned}$$

b

$$\begin{aligned} \sum_{1 \leq i<j<k \leq 4} a_{i j k} &= \sum_{k=1}^4 \sum_{j=1}^{k-1} \sum_{i=1}^{j-1} a_{ijk}\\ &= a_{123} +(a_{124} + (a_{134} + a_{234})) \end{aligned}$$

5

将$j$替换为$k$，导致两个指标相同。

6

$$\begin{aligned} \sum_{k}[1 \le j \le k \le n] &= \sum_{k}[1 \le j \le n] [ j \le k \le n] \\ &= [1 \le j \le n]\sum_{j \le k \le n} 1\\ &= [1 \le j \le n](n - j + 1) \end{aligned}$$

7

$$\begin{aligned} x^{\overline{m}} &= x(x+1)\ldots(x+m - 1)\\ \nabla(x^{\overline{m}} ) &= x^{\overline{m}}-(x-1)^{\overline{m}}\\ &= x(x+1)\ldots(x+m - 1)- (x-1)x\ldots(x+m - 2)\\ &= x(x+1)\ldots(x+m - 2)(x+m - 1- x +1)\\ &= mx(x+1)\ldots(x+m - 2)\\ &= m x^{\overline{m-1}} \end{aligned}$$

8

当$m> 0$时，

$$\begin{aligned} 0^{\underline m} &= 0\times (0-1)\times \ldots (0- (m - 1))\\ &= 0 \end{aligned}$$

当$m\le 0$时，

$$\begin{aligned} 0^{\underline m} &= (0+1)^{-1}\times \ldots (0+m)^{-1}\\ &= \frac 1{|m|!} \end{aligned}$$

9

$$\begin{aligned} x^{\overline{m+n}} &= x\times \ldots\times (x+m -1)\times (x+m)\times \ldots \times (x+m+n-1)\\ &=x^{\overline m} (x+m)^{\overline {n}} \end{aligned}$$

取$m=-n$，那么

$$\begin{aligned} x^{\overline{-n+n}} &= x^{\overline 0}\\ &=1\\ &=x^{\overline {-n}} (x-n)^{\overline {n}}\\ x^{\overline {-n}}&= \frac{1}{(x-n)^{\overline {n}}}\\ &=\frac 1 {(x-n)(x-n+1)\ldots (x-1)}\\ &= \frac 1 {(x-1)^{\underline n}} \end{aligned}$$

10

显然有

$$\Delta(uv)=\Delta (vu)$$

所以只要证明

$$u \Delta v+\mathrm{E} v \Delta u= v \Delta u+\mathrm{E} u \Delta v$$

即可。

左边：

$$\begin{aligned} u \Delta v+\mathrm{E} v \Delta u &= u(x)(v(x+1)-v(x))+v(x+1)(u(x+1)-u(x))\\ &= u(x)v(x+1)-u(x)v(x)+v(x+1)u(x+1) -u(x)v(x+1)\\ &=u(x+1)v(x+1) -u(x)v(x) \end{aligned}$$

右边：

$$\begin{aligned} v \Delta u+\mathrm{E} u \Delta v &= v(x)(u(x+1)-u(x))+u(x+1)(v(x+1)-v(x))\\ &= v(x)u(x+1)-v(x)u(x)+u(x+1)v(x+1) -u(x+1)v(x)\\ &=u(x+1)v(x+1) -u(x)v(x) \end{aligned}$$

因此

$$u \Delta v+\mathrm{E} v \Delta u= v \Delta u+\mathrm{E} u \Delta v$$

基础题

11

$$\begin{aligned} \sum_{0 \le k<n}\left(a_{k+1}-a_{k}\right) b_{k} &= \sum_{0 \le k<n}a_{k+1}b_{k}-\sum_{0 \le k<n}a_{k}b_{k} \\ &= \sum_{0 \le k<n}a_{k+1}b_{k}-\sum_{0 \le k< n}a_{k+1}b_{k+1}-a_0b_0 + a_n b_n \\ &=a_n b_n -a_0b_0 -\sum_{0 \le k<n}a_{k+1}(b_{k+1}- b_k) \end{aligned}$$

12

注意到

$$\begin{aligned} p(2k) & = 2k+c\\ p(2k+1) &= 2k+1 -c \end{aligned}$$

如果$c$是奇数，那么

$$\begin{aligned} \big \{2k | k\in \mathbb Z \big\} & = \big \{2k+1-c | k\in \mathbb Z \big\}\\ \big \{2k+1 | k\in \mathbb Z \big\} & = \big \{2k+c | k\in \mathbb Z \big\} \end{aligned}$$

如果$c$是偶数，那么

$$\begin{aligned} \big \{2k | k\in \mathbb Z \big\} & = \big \{2k+c | k\in \mathbb Z \big\} \\ \big \{2k+1 | k\in \mathbb Z \big\} & = \big \{2k+1-c | k\in \mathbb Z \big\} \end{aligned}$$

因此

$$\begin{aligned} \mathbb Z&= \big \{2k | k\in \mathbb Z \big\}\bigcup \{2k+1| k\in \mathbb Z \}\\ &= \big \{2k+c | k\in \mathbb Z \big\} \bigcup \big \{2k+1-c | k\in \mathbb Z \big\} \end{aligned}$$

13

记

$$R_n = \sum_{k=0}^{2n}(-1)^{k} k^{2}$$

那么

$$\begin{aligned} R_{0}&=0 \\ R_n&= R_{n-1} -(2n-1)^2 + (2n)^2\\ &= R_{n-1} +4n-1 \end{aligned}$$

递归式

$$\begin{aligned} R_{0}&=\alpha \\ R_{n}&=R_{n-1}+\beta+\gamma n, \quad n>0 \end{aligned}$$

解的一般形式为

$$R_{n}=A(n) \alpha+B(n) \beta+C(n) \gamma$$

根据22页的结果可得

$$\begin{aligned} A(n)&=1\\ B(n)&=n\\ C(n)&= \frac{n(n+1)}{2} \end{aligned}$$

此处

$$\begin{aligned} \alpha&=0\\ \beta&=-1\\ \gamma&= 4 \end{aligned}$$

所以

$$\begin{aligned} R_n &= -n+2n(n+1)\\ &=2n^2 +n\\ \sum_{k=0}^{2n}(-1)^{k} k^{2}&= 2n^2 +n\\ &=n(2n+1) \\ &=\frac{2n(2n+1)}{2} \\ \sum_{k=0}^{2n+1}(-1)^{k} k^{2} &= \sum_{k=0}^{2n}(-1)^{k} k^{2}-(2n+1)^2\\ &= 2n^2 +n -(2n+1)^2\\ &= -2n^2-3n-1\\ &=-(2n+1)(n+1)\\ &= -\frac{(2n+1)(2n+2)}{2} \end{aligned}$$

因此

$$\sum_{k=0}^{n}(-1)^{k} k^{2} =(-1)^n \frac{n(n+1)}{2}$$

注意该方法有特殊性，更一般的方法应该是令

$$R_n =\sum_{k=0}^{n}(-1)^{k} k^{2}$$

然后得到递推式

$$R_{n}=R_{n-1}+(-1)^{n}\left(\beta+\gamma n + \delta n^{2}\right)$$

14

$$\begin{aligned} \sum_{k=1}^{n} k 2^{k} &= \sum_{1 \le j \le k \le n} 2^{k}\\ &=\sum_{1 \le j\le n}\sum_{j\le k \le n} 2^{k} \\ &= \sum_{1 \le j\le n} (2^{n+1} - 2^j)\\ &= \sum_{1 \le j\le n}2^{n+1}- \sum_{1 \le j\le n}2^j\\ &= n2^{n+1}- (2^{n+1}- 2)\\ &=(n-1)2^{n+1} +2 \end{aligned}$$

15

记

$$\begin{aligned} S_n &= \sum_{k=1}^n k^2\\ T_n &= \sum_{k=1}^n k^3 \end{aligned}$$

注意到

$$\begin{aligned} S_n +T_n&= \sum_{k=1}^n k^2+\sum_{k=1}^n k^3\\ &= \sum_{k=1}^nk(k+1)k \\ &=2 \sum_{1 \le j \le k \le n} j k \end{aligned}$$

回顾31页2.33

$$\sum_{1 \le j \le k \le n} a_{j} a_{k}=\frac{1}{2}\left(\left(\sum_{k=1}^{n} a_{k}\right)^{2}+\sum_{k=1}^{n} a_{k}^{2}\right)$$

因此

$$\begin{aligned} S_n +T_n &=2 \sum_{1 \le j \le k \le n} j k\\ &= \left(\sum_{k=1}^{n} k\right)^{2}+\sum_{k=1}^{n} k^{2}\\ &=\left(\sum_{k=1}^{n} k\right)^{2}+T_n\\ S_n&= \sum_{k=1}^n k^3\\ &=\left(\sum_{k=1}^{n} k\right)^{2} \end{aligned}$$

16

$$\begin{aligned} x^{\underline m} (x-m)^{\underline n} &=x(x-1)\ldots(x-m+1) \times (x-m)(x-m-1)\ldots(x-m-n+1)\\ &= x^{\underline {(m+n)}} \\ x^{\underline n} (x-n)^{\underline m} &=x(x-1)\ldots(x-n+1) \times (x-n)(x-n-1)\ldots(x-n-m+1)\\ &= x^{\underline {(m+n)}}\\ x^{\underline m} (x-m)^{\underline n} &=x^{\underline n} (x-n)^{\underline m}\\ x^{\underline m} /(x-n)^{\underline m} &= x^{\underline n} /(x-m)^{\underline n} \end{aligned}$$

17

第一个等式：

$$\begin{aligned} x^{\overline m} &= x(x+1)\ldots(x+m-1)\\ &= (x+m-1)^{\underline m}\\ x^{\overline m} &= x(x+1)\ldots(x+m-1)\\ &=(-1)^m(-x-m+1)(-x-m+2) \ldots(-x-1)(-x)\\ &=(-1)^{m}(-x)^{\underline m}\\ 1/(x-1)^{\underline{-m}} &= x(x+1)\ldots(x+m-1)\\ &= x^{\overline m} \end{aligned}$$

第二个等式：

$$\begin{aligned} x^{\underline m} &= x(x-1)\ldots(x-m+1)\\ &= (x-m+1)^{\overline m}\\ x^{\underline m} &= x(x-1)\ldots(x-m+1)\\ &=(-1)^m(-x)(-x+1) \ldots(-x+m-1)\\ &=(-1)^{m}(-x)^{\overline m}\\ 1/(x+1)^{\overline{-m}} &= x(x-1)\ldots(x-m+1)\\ &= x^{\underline m} \end{aligned}$$

18

注意到

$$\begin{aligned} |\Re a_{k} | &\le |a_k| \\ |\Im a_{k} | &\le |a_k| \\ |a_k| &\le |\Re a_{k} |+|\Im a_{k} | \end{aligned}$$

$\Rightarrow$:

因为$\sum_{k \in K} a_{k}$绝对收敛，所以$\sum_{k \in K}\Re a_{k},\sum_{k \in K}\Im a_{k}$绝对收敛，因此存在$B_1,B_2$，使得对所有$F\subseteq K$，都有

$$\begin{aligned} \sum_{k \in F}|\Re a_{k}|& \le B_1\\ \sum_{k \in F}|\Im a_{k}| & \le B_2 \end{aligned}$$

记$B=B_1+B_2$，那么

$$\begin{aligned} \sum_{k \in F}| a_{k}| &\le \sum_{k \in F}(|\Re a_{k}|+|\Im a_{k}| )\\ &\le \sum_{k \in F}|\Re a_{k}|+\sum_{k \in F}|\Im a_{k}| \\ & \le B_1+ B_2\\ &\le B \end{aligned}$$

$\Leftarrow$:

对所有$F\subseteq K$，都有

$$\begin{aligned} \sum_{k \in F}|\Re a_{k}| &\le \sum_{k \in F}|a_{k}|\le B\\ \sum_{k \in F}|\Im a_{k}| &\le \sum_{k \in F}|a_{k}|\le B \end{aligned}$$

所以$\sum_{k \in K}\Re a_{k},\sum_{k \in K}\Im a_{k}$绝对收敛，因此$\sum_{k \in K} a_{k}$绝对收敛。

19

$$\begin{aligned} T_{0}&=5 \\ 2 T_{n}&=n T_{n-1}+3 \times n ! \end{aligned}$$

回顾38页：

$$\begin{aligned} a_{n}&= 2\\ b_n &= n\\ s_{n} &=\frac{a_{n-1} a_{n-2} \cdots a_{1}}{b_{n} b_{n-1} \cdots b_{2}}\\ &= \frac{2^{n-1}}{n!} \end{aligned}$$

可得

$$\begin{aligned} \frac{2^nT_{n}}{n !}&=\frac{2^{n-1}T_{n-1}}{(n-1)!}+3 \times 2^{n-1}\\ \frac{2^nT_{n}}{n !}&= \frac{2^{0}T_{0}}{0!} +\sum_{k=0}^{n-1} 3\times 2^{k-1}\\ \frac{2^nT_{n}}{n !}&= 5+3(2^n -1)\\ T_n&= \frac{(3\times 2^n +2) n!}{2^n}\\ &= 3n! +\frac{n!}{2^{n-1}} \end{aligned}$$

20

定义

$$S_n = \sum_{k=0}^{n} k H_{k}$$

那么

$$\begin{aligned} S_n + (n+1)H_{n+1} &= \sum_{k=1}^{n+1} k H_{k}\\ &= \sum_{k=0}^{n} (k+1) H_{k+1}\\ &=\sum_{k=0}^{n} kH_{k+1} +\sum_{k=0}^{n} H_{k+1}\\ &= \sum_{k=0}^{n} k\left( H_{k} +\frac 1 {k+1} \right) +\sum_{k=1}^{n+1} H_{k}\\ &= \sum_{k=0}^{n} k H_{k} + \sum_{k=0}^{n} \frac{k}{k+1} +\sum_{k=0}^{n+1} H_{k}\\ &= S_n + \sum_{k=0}^{n} \left(1- \frac{1}{k+1}\right)+ \sum_{k=0}^{n+1} H_{k}\\ &= S_n + n+1-H_{n+1}+ \sum_{k=0}^{n+1} H_{k}\\ &= S_n + n+1+ \sum_{k=0}^{n} H_{k}\\ \sum_{k=0}^{n} H_{k}&= (n+1)H_{n+1}- (n+1)\\ \sum_{k=0}^{n-1} H_{k}&= n H_{n}-n \end{aligned}$$

21

和式一：

$$\begin{aligned} S_{n+1}&=\sum_{k=0}^{n+1}(-1)^{n+1-k}\\ &=(-1)^{n+1}+\sum_{k=1}^{n+1}(-1)^{n+1-k}\\ &=(-1)^{n+1}+\sum_{k=1}^{n+1}(-1)^{n+1-k}\\ &=(-1)^{n+1}+\sum_{k=0}^{n}(-1)^{n-k}\\ &= (-1)^{n+1} + S_n\\ -S_n +1&=\sum_{k=0}^{n}(-1)^{n+1-k} + (-1)^{n+1-n-1}\\ &=\sum_{k=0}^{n+1}(-1)^{n+1-k} \\ &= S_{n+1}\\ -S_n +1&= (-1)^{n+1} + S_n\\ S_n &=\frac{1-(-1)^{n+1}}{2}\\ &=\frac{1+(-1)^{n}}{2} \end{aligned}$$

和式二：

$$\begin{aligned} T_{n+1}&=\sum_{k=0}^{n+1}(-1)^{n+1-k}k\\ &=\sum_{k=1}^{n+1}(-1)^{n+1-k}k\\ &= \sum_{k=0}^{n}(-1)^{n-k}(k+1)\\ &= \sum_{k=0}^{n}(-1)^{n-k}k+\sum_{k=0}^{n}(-1)^{n-k}\\ &= T_n +S_n\\ -T_n +n+1&=\sum_{k=0}^{n}(-1)^{n+1-k}k + (-1)^{n+1-n-1}(n+1)\\ &=\sum_{k=0}^{n+1}(-1)^{n+1-k}k \\ &= T_{n+1}\\ -T_n +n+1&=T_n +S_n\\ T_n&=\frac{n+1- S_n}{2} \end{aligned}$$

和式三：

$$\begin{aligned} U_{n+1}&=\sum_{k=0}^{n+1}(-1)^{n+1-k}k^2\\ &=\sum_{k=1}^{n+1}(-1)^{n+1-k}k^2\\ &= \sum_{k=0}^{n}(-1)^{n-k}(k+1)^2\\ &= \sum_{k=0}^{n}(-1)^{n-k}k^2+2\sum_{k=0}^{n}(-1)^{n-k} k +\sum_{k=0}^{n}(-1)^{n-k}\\ &= U_n +2T_n +S_n\\ -U_n +(n+1)^2&=\sum_{k=0}^{n}(-1)^{n+1-k}k^2 + (-1)^{n+1-n-1}(n+1)^2\\ &=\sum_{k=0}^{n+1}(-1)^{n+1-k}(k+1)^2 \\ &= U_{n+1}\\ -U_n +(n+1)^2&=U_n +2T_n +S_n\\ U_n&=\frac{(n+1)^2-2T_n- S_n}{2}\\ &= \frac{(n+1)^2-(n+1)+S_n-S_n}{2}\\ &=\frac{(n+1)n}{2} \end{aligned}$$

22

记：

$$\begin{aligned} S_1&=\sum_{1 \le j<k \le n}\left(a_{j} b_{k}-a_{k} b_{j}\right)\left(A_{j} B_{k}-A_{k} B_{j}\right)\\ S_2&=\sum_{1 \le k<j \le n}\left(a_{k} b_{j}-a_{j} b_{k}\right)\left(A_{k} B_{j}-A_{j} B_{k}\right)\\ &=\sum_{1 \le j<k \le n}\left(a_{j} b_{k}-a_{k} b_{j}\right)\left(A_{j} B_{k}-A_{k} B_{j}\right) \end{aligned}$$

将$j,k$互换，我们得到

$$\begin{aligned} S_1= S_2 \end{aligned}$$

另一方面

$$\begin{aligned} 2S_1 &=S_1+ S_2\\ &=\sum_{1 \le j ,k \le n}\left(a_{j} b_{k}-a_{k} b_{j}\right)\left(A_{j} B_{k}-A_{k} B_{j}\right)\\ &=\sum_{1 \le j ,k \le n}\left( a_jA_j b_k B_k - b_j A_ja_kB_k -a_jB_jb_kA_k + b_j B_ja_k A_k\right)\\ &= \sum_{k=1}^n \sum_{j=1}^{n} a_jA_j b_k B_k -\sum_{k=1}^n \sum_{j=1}^{n} b_j A_ja_kB_k -\sum_{k=1}^n \sum_{j=1}^{n} a_jB_jb_kA_k +\sum_{k=1}^n \sum_{j=1}^{n} b_j B_ja_k A_k\\ &=\left(\sum_{k=1}^n b_k B_k \right)\left( \sum_{j=1}^n a_jA_j\right) -\left(\sum_{k=1}^n a_kB_k\right) \left(\sum_{j=1}^n b_jA_j\right)- \left(\sum_{k=1}^n b_kA_k \right)\left(\sum_{j=1}^n a_jB_j\right)+ \left(\sum_{k=1}^n a_kA_k \right)\left(\sum_{j=1}^n b_jB_j\right)\\ &=2 \left( \sum_{k=1}^n a_kA_k\right)\left(\sum_{k=1}^n b_k B_k \right) -2\left(\sum_{k=1}^n a_kB_k\right) \left(\sum_{k=1}^n b_kA_k\right)\\ S_1&=\left( \sum_{k=1}^n a_kA_k\right)\left(\sum_{k=1}^n b_k B_k \right)- \left(\sum_{k=1}^n a_kB_k\right) \left(\sum_{k=1}^n b_kA_k\right) \end{aligned}$$

因此

$$\sum_{1 \le j<k \le n}\left(a_{j} b_{k}-a_{k} b_{j}\right)\left(A_{j} B_{k}-A_{k} B_{j}\right)=\left( \sum_{k=1}^n a_kA_k\right)\left(\sum_{k=1}^n b_k B_k \right)- \left(\sum_{k=1}^n a_kB_k\right) \left(\sum_{k=1}^n b_kA_k\right)$$

特别的，取

$$\begin{aligned} A_j &= a_j\\ B_j &= b_j \end{aligned}$$

可得

$$\sum_{1 \le j<k \le n}\left(a_{j} b_{k}-a_{k} b_{j}\right)^{2}=\left(\sum_{k=1}^{n} a_{k}^{2}\right)\left(\sum_{k=1}^{n} b_{k}^{2}\right)-\left(\sum_{k=1}^{n} a_{k} b_{k}\right)^{2}$$

23

解法1：

$$\begin{aligned} \sum_{k=1}^{n}\frac{2 k+1}{k(k+1)} &= \sum_{k=1}^{n}\left(\frac 1 k + \frac 1{k+1}\right)\\ &=\sum_{k=1}^{n} \frac 1 k + \sum_{k=1}^{n} \frac 1{k+1}\\ &= H_n + H_{n+1}-1 \end{aligned}$$

解法2：

回顾分部求和法：

$$\sum_{k=1}^n u(k) \Delta v(k)=u(n+1) v(n+1)-u(1)v(1)-\sum_{k=1}^n \mathrm{E} v(k) \Delta u(k)$$

这里取

$$\begin{aligned} u(k)& = 2k+1\\ v(k)&= -\frac1 {k}\\ u(k)\Delta v(k) &= (2k+1)\left(-\frac 1 {k+1} +\frac 1 k\right)\\ &=\frac{2k+1}{k(k+1)} \end{aligned}$$

因此

$$\begin{aligned} \sum_{k=1}^{n}\frac{2 k+1}{k(k+1)} &=\sum_{k=1}^n u(k) \Delta v(k)\\ &=u(n+1) v(n+1)-u(1)v(1)-\sum_{k=1}^n \mathrm{E} v(k) \Delta u(k)\\ &= -\frac{2n+3}{n+1} + 3+\sum_{k=1}^n \frac 1 {k+1} 2\\ &= 1 -\frac 1 {n+1} + 2 \left( H_{n+1} - 1\right)\\ &= H_n + H_{n+1}-1 \end{aligned}$$

24

回顾分部求和法：

$$\sum_{k=1}^n u(k) \Delta v(k)=u(n+1) v(n+1)-u(1)v(1)-\sum_{k=1}^n \mathrm{E} v(k) \Delta u(k)$$

这里取

$$\begin{aligned} u(k)& = H(k)\\ v(k)&= -\frac1 {k+1}\\ u(k)\Delta v(k) &= H(k)\left(\frac 1 {k+1}-\frac 1 {k+2}\right)\\ &=\frac{H(k)}{(k+1)(k+2)} \end{aligned}$$

因此

$$\begin{aligned} \sum_{k=1}^{n -1}\frac{H(k)}{(k+1)(k+2)} &=\sum_{k=1}^{n-1} u(k) \Delta v(k)\\ &=u(n) v(n)-u(1)v(1)-\sum_{k=1}^{n-1} \mathrm{E} v(k) \Delta u(k)\\ &= -\frac{H(n)}{n+1} + \frac 12 +\sum_{k=1}^{n-1} \frac 1 {k+2} \frac 1 {k+1}\\ &= -\frac{H(n)}{n+1} + \frac 12 +\sum_{k=1}^{n-1} \left(\frac 1 {k+1}-\frac 1 {k+2} \right)\\ &= -\frac{H(n)}{n+1} + \frac 12+\frac 1 2 -\frac 1 {n+1}\\ &= 1-\frac{H(n)}{n+1}-\frac 1 {n+1} \end{aligned}$$

25

对应关系：

$$\begin{aligned} \sum_{k \in K} c a_{k}=c \sum_{k \in K} a_{k} &\leftrightarrow \prod_{k \in K} a_{k}^c= \left(\sum_{k \in K} a_{k} \right)^c\\ \sum_{k \in K}\left(a_{k}+b_{k}\right)=\sum_{k \in K} a_{k}+\sum_{k \in K} b_{k} &\leftrightarrow \prod_{k \in K}a_{k}b_{k}=\prod_{k \in K} a_{k}\prod_{k \in K} b_{k}\\ \sum_{k \in K} a_{k}=\sum_{p(k) \in K} a_{p(k)}&\leftrightarrow \prod_{k \in K} a_{k}=\prod_{p(k) \in K} a_{p(k)}\\ \sum_{P(j, k)} a_{j, k}=\sum_{j, k} a_{j, k}[P(j, k)]&\leftrightarrow \prod_{P(j, k)} a_{j, k}=\prod_{j, k} a_{j, k}^{[P(j, k)]}\\ \sum_{j \in J} \sum_{k \in K} a_{j, k}=\sum_{k \in K} \sum_{k \in K} a_{j, k}&\leftrightarrow \prod_{j \in J} \prod_{k \in K} a_{j, k}=\prod_{k \in K} \prod_{k \in K} a_{j, k} \end{aligned}$$

性质的改变基本符合如下规律：

• $\sum \to \prod$
• 加法变乘法。
• 乘法变乘方。

26

记

$$S =\prod_{k=1}^{n} a_{k}$$

注意到

$$\begin{aligned} P=\prod_{1 \leq j \le k \le n} a_{j} a_{k} \end{aligned}$$

交换$j,k$可得

$$\begin{aligned} P &=\prod_{1 \leq k \le j \le n} a_{j} a_{k} \end{aligned}$$

因此

$$\begin{aligned} P^2 &=\prod_{1 \leq j \le k \le n} a_{j} a_{k} \prod_{1 \leq k \le j \le n} a_{j} a_{k}\\ &= \left(\prod_{1 \leq j,k \le n} a_{j} a_{k} \right) \times \left(\prod_{1 \leq k = j \le n} a_{k}^2 \right) \\ &= \left(\prod_{j=1}^n a_{j} \right) \times \left(\prod_{k=1}^n a_{k} \right) \times \left(\prod_{k=1}^n a_{k} \right)^{2n}\\ P&= \left(\prod_{k=1}^n a_{k} \right)^{n+1}\\ &= S^{n+1} \end{aligned}$$

27

$$\begin{aligned} \Delta\left(c^{\underline{x}}\right) &= c(c-1)\ldots (c-x+1)(c-x)-c(c-1)\ldots (c-x+1)\\ &=c(c-1)\ldots (c-x+1) (c-x-1)\\ &= \frac{c^{\underline{x+2}}}{c-x} \end{aligned}$$

所以

$$\begin{aligned} \Delta\left(c^{\underline{x+c}}\right) &= \frac{c^{\underline{x+c+2}}}{c-x-c}\\ &= -\frac{c^{\underline{x+c+2}}}{x} \end{aligned}$$

取$c=-2$可得

$$\begin{aligned} \Delta\left((-2)^{\underline{x-2}}\right) &= -\frac{(-2)^{\underline{x}}}{x} \end{aligned}$$

因此

$$\begin{aligned} \sum_{k=1}^{n}\frac{(-2)^{k}} { k} &=\sum_{k=1}^n \Delta\left(-(-2)^{\underline{k-2}}\right)\\ &= -(-2)^{\underline{n-1}} + (-2)^{-1}\\ &= -(-2)(-3)\times \ldots \times (-2 -n+2) +\frac 1 {-2+1}\\ &=(-1)^n n ! -1 \end{aligned}$$

28

第二个等号位置已经有错误了，因为分解的两个级数不绝对收敛。

29

$$\begin{aligned} \sum_{k=1}^{n}\frac{(-1)^{k} k} {4 k^{2}-1} &= \sum_{k=1}^{n}\frac{(-1)^{k} k} {(2k-1)(2k+1)}\\ &= \frac 1 4\sum_{k=1}^{n}(-1)^{k} \left(\frac 1 {2k-1} + \frac 1 {2k+1}\right)\\ &= \frac 1 4\sum_{k=1}^{n}\left(\frac {(-1)^k} {2k-1} -\frac {(-1)^{k+1}} {2k+1} \right)\\ &= \frac 1 4 \left(-1 -\frac {(-1)^{n+1}} {2n+1}\right)\\ &=-\frac1 4 + \frac {(-1)^{n}} {8n+4} \end{aligned}$$

30

记目标为$n$，假设其可表示为$k$个整数和：

$$\begin{aligned} n &= \sum_{j=1}^k (i+j)\\ &= \frac{k(2i + k+ 1)}{2}\\ 2n &= k(2i+ k + 1) \end{aligned}$$

注意这里$k$和$2i+k+1$奇偶性不同。

这里$n=1050$，因此

$$k(2i+ k + 1) = 2^2 \times 3 \times 5^2 \times 7$$

奇数因子数量为$2\times 3\times 2=12$，而$k< 2i+k+1$，所以数量为$12$。

31

等式一：

$$\begin{aligned} \sum_{k \ge 2}(\zeta(k)-1) &= \sum_{k \ge 2}\sum_{j \ge 2} \frac{1}{j^{k}}\\ &=\sum_{j \ge 2} \sum_{k \ge 2} \frac{1}{j^{k}}\\ &= \sum_{j \ge 2} \frac{\frac 1 {j^2}}{1-\frac 1 j}\\ &= \sum_{j \ge 2} \frac{1}{(j- 1)j}\\ &= \sum_{j \ge 2} \left(\frac 1 {j-1} - \frac 1 j \right)\\ &= 1 \end{aligned}$$

等式二：

$$\begin{aligned} \sum_{k \ge 1}(\zeta(2k)-1) &= \sum_{k \ge 1}\sum_{j \ge 2} \frac{1}{j^{2k}}\\ &=\sum_{j \ge 2} \sum_{k \ge 1} \frac{1}{j^{2k}}\\ &= \sum_{j \ge 2} \frac{\frac 1 {j^2}}{1-\frac 1 {j^2}}\\ &= \sum_{j \ge 2} \frac{1}{j^2 - 1}\\ &=\frac 1 2 \sum_{j \ge 2} \left(\frac 1 {j-1} - \frac 1 {j+1} \right)\\ &=\frac 1 2 \left(1 + \frac 1 2 \right)\\ &=\frac 3 4 \end{aligned}$$

32

左边：

$$\begin{aligned} \sum_{k \ge 0} \min (k, x \dot - k) &= \sum_{0\le k \le x} \min (k, x-k) + \sum_{k>x} \min (k, 0) \\ &= \sum_{0\le k \le \lfloor x \rfloor} \min (k, x-k)\\ &= \sum_{0\le k \le \lfloor x/ 2 \rfloor} k + \sum_{\lfloor x/ 2 \rfloor +1 \le k \le \lfloor x \rfloor} (x-k)\\ &= \frac{(\lfloor x/ 2 \rfloor + 1)\lfloor x/ 2 \rfloor}{2} + (\lfloor x \rfloor-\lfloor x/ 2 \rfloor ) x -\frac{(\lfloor x \rfloor+\lfloor x/ 2 \rfloor +1 )(\lfloor x \rfloor-\lfloor x/ 2 \rfloor )}{2}\\ &= (\lfloor x /2 \rfloor+ 1)\lfloor x /2 \rfloor-\frac{(\lfloor x \rfloor+ 1)\lfloor x \rfloor}{2} + (\lfloor x \rfloor-\lfloor x/ 2 \rfloor ) x \end{aligned}$$

如果$2i \le x < 2i+1, i\in \mathbb Z$，那么

$$\begin{aligned} \sum_{k \ge 0} \min (k, x \dot - k) &= (\lfloor x /2 \rfloor+ 1)\lfloor x /2 \rfloor-\frac{(\lfloor x \rfloor+ 1)\lfloor x \rfloor}{2} + (\lfloor x \rfloor-\lfloor x/ 2 \rfloor ) x\\ &= (i+1)i - \frac{(2i+1)2i}{2} + (2i- i) x \\ &= ix-i^2 \end{aligned}$$

如果$2i -1\le x < 2i,i-1/2\le x/ 2 < i,i\in \mathbb Z$，那么

$$\begin{aligned} \sum_{k \ge 0} \min (k, x \dot - k) &= (\lfloor x /2 \rfloor+ 1)\lfloor x /2 \rfloor-\frac{(\lfloor x \rfloor+ 1)\lfloor x \rfloor}{2} + (\lfloor x \rfloor-\lfloor x/ 2 \rfloor ) x\\ &=i(i-1) -\frac{2i(2i-1)}{2} + (2i-1- (i-1)) x \\ &= ix -i^2 \end{aligned}$$

右边：

$$\begin{aligned} \sum_{k \ge 0}(x \dot - (2 k+1)) &= \sum_{0\le 2k + 1 \le x} (x-2k-1) + \sum_{2k + 1>x} 0 \\ &= \sum_{0\le k\le \lfloor (x-1)/2 \rfloor} (x-2k-1)\\ &= x( \lfloor (x-1)/2 \rfloor + 1) - \frac{(2+ 2\lfloor (x-1)/2 \rfloor )(\lfloor (x-1)/2 \rfloor +1)}{2} \end{aligned}$$

如果$2i \le x < 2i+1,i -1/2\le (x-1)/2 <i, i\in \mathbb Z$，那么

$$\begin{aligned} \sum_{k \ge 0}(x \dot - (2 k+1)) &= x( \lfloor (x-1)/2 \rfloor + 1) - \frac{(2+ 2\lfloor (x-1)/2 \rfloor )(\lfloor (x-1)/2 \rfloor +1)}{2}\\ &= x (i-1 + 1) - \frac{(2+ 2(i-1))(i-1 +1)}{2}\\ &= ix- i^2 \end{aligned}$$

如果$2i-1 \le x < 2i,i -1\le (x-1)/2 <i -1/2, i\in \mathbb Z$，那么

$$\begin{aligned} \sum_{k \ge 0}(x \dot - (2 k+1)) &= x( \lfloor (x-1)/2 \rfloor + 1) - \frac{(2+ 2\lfloor (x-1)/2 \rfloor )(\lfloor (x-1)/2 \rfloor +1)}{2}\\ &= x (i-1 + 1) - \frac{(2+ 2(i-1))(i-1 +1)}{2}\\ &= ix- i^2 \end{aligned}$$

因此

$$\sum_{k \ge 0} \min (k, x \dot - k) = \sum_{k \ge 0}(x \dot - (2 k+1))$$

重点

27, 30, 32