$0$

\left\{ \begin{aligned} 0 & = & x=0 \\ x & = & x>0 \\ -x & = & x<0 \end{aligned} \right. =>|x|

$\sum_{0\leq k\leq 5}^{}a_k=a_0 + a_1 + a_2 + a_3 + a_4 + a_5$
$\sum_{0\leq k^2\leq 5}^{}a_k=a_4 + a_1 + a_0 + a_1 + a_4$

$$(\sum_{j=1}^{n}a_j)(\sum_{k=1}^{n}\frac{1}{a_k})=\sum_{j=1}^{n}\sum_{k=1}^{n}\frac{a_j}{a_k}=\sum_{k=1}^{n}\sum_{k=1}^{n}\frac{a_k}{a_k}=\sum_{k=1}^{n}=n^2$$

$\sum_{k}^{}[1\leq j\leq k\leq n]=\sum_{j\leq k\leq n}^{}[1\leq j\leq n]=[1\leq j\leq n](n-j+1)$

$$\nabla(x^{\overline m})=x^{\overline m}-(x-1)^{\overline m}\\=x(x+1)...(x+m-1)-(x-1)x(x+1)...(x+m-2)\\=x(x+1)...(x+m-2)((x+m-1)-(x-1))\\=x(x+1)...(x+m-2)(x-x+m-1+1)\\=mx(x+1)...(x+m-2)\\=mx^{\overline{m-1}}$$

$$x^{\overline{m+n}}=x(x+1)...(x+m+n-1)\\=x(x+1)...(x+m-1)(x+m)...(x+m+n-1)\\=x^{\overline m}(x+m)^{\overline n}$$
$x^{\overline {-1}}=\frac{1}{x-1},x^{\overline {-2}}=\frac{1}{(x-1)(x-2)},x^{\overline{-n}}=\frac{1}{(x-1)...(x-n)},n>0$

$$\Delta(uv)=u\Delta v+Ev\Delta u$$

$$\Delta(u(x)v(x))=u(x+1)v(x+1)-u(x)v(x)\\=u(x+1)v(x+1)-u(x+1)v(x)+u(x+1)v(x)-u(x)v(x)\\=u(x+1)(v(x+1)-v(x))+v(x)(u(x+1)-u(x))\\=u(x+1)\Delta v(x)+v(x)\Delta u(x)\\=Eu(x)\Delta v(x)+v(x)\Delta u(x)$$

$$\sum_{0\leq k<n}^{}(a_{k+1}-a_k)b_k=a_nb_n-a_0b_0-\sum_{0\leq k<n}^{}a_{k+1}(b_{k+1}-b_k),\quad n\geq 0$$

$$\sum_{0\leq k<n}^{}(a_{k+1}-a_k)b_k=\sum_{0\leq k<n}^{}a_{k+1}b_{k}-a_kb_k\\=\sum_{0\leq k<n}^{}a_{k+1}b_k-\sum_{0\leq k<n}^{}a_kb_k\\=\sum_{0\leq k<n}^{}-[(\sum_{0\leq k<n}^{}a_{k+1}b_{k+1})+a_0b_0-a_nb_n]\\=(\sum_{0\leq k<n}^{}a_{k+1}b_k-a_{k+1}b_{k+1})-a_0b_0+a_nb_n\\=\sum_{0\leq k<n}^{}a_nb_n-a_0b_0+\sum_{0\leq k<n}^{}a_{k+1}(b_k-b_{k+1})\\=a_nb_n-a_0b_0-\sum_{0\leq k<n}^{}a_{k+1}(b_{k+1}-b_k)$$

$$R_0=\alpha\\R_n=R_{n-1}+(-1)^n(\beta +\gamma n+\delta n^2),n\geq 1$$

$$\sum_{k=1}^{n}k2^k=\sum_{1\leq j\leq k\leq n}^{}2^k=\sum_{1\leq j\leq n}^{}\sum_{j\leq k\leq n}^{}2^k\\=\sum_{1\leq j\leq n}^{}(\frac{2(1-2^n)}{1-2}-\frac{2(1-2^{j-1})}{1-2})\\=\sum_{1\leq j\leq n}^{}(2^{n+1}-2^j)=(\sum_{1\leq j\leq n}^{}2^{n+1})-\sum_{1\leq j\leq n}^{}2^j\\=n2^{n+1}-\frac{2(1-2^n)}{1-2}=(n-1)2^{n+1}+2$$

$$\sum_{1\leq k\leq n}^{}k^3=(2\sum_{1\leq j\leq k\leq n}^{}jk)-\sum_{1\leq k\leq n}^{}k^2\\=(\sum_{k=1}^{n}k)^2+(\sum_{k=1}^{n}k^2)-\sum_{1\leq k\leq n}^{}k^2\\=(\frac{n(n+1)}{2})^2=\frac{n^2(n+1)^2}{4}$$

$$x^{\overline m}=(-1)^m(-x)^{\underline m}=(x+m-1)^{\underline m}=\frac{1}{(x-1)^{\underline {-m}}}$$

$$x^{m}\underline =(-1)^m(-x)^{\overline m}=(x-m+1)^{\overline m}=\frac{1}{(x-1)^{\overline {-m}}}$$

（习题九的答案给出了$x^{\overline {-m}}$的定义。）

$x^\overline{m}=x(x+1)...(x+m-1)=(x+m-1)...(x+1)x=(x+m-1)^\underline m$

$(-1)^m(-x^\underline m)=(-1)^m(-x)(-x-1)...(-x-m+1)=(-1)^{2m}x(x+1)...(x+m-1)=x^\overline m$

$\frac{1}{(x-1)^\underline{-m}}=\frac{1}{\frac{1}{(x-1+1)(x-1+2)...(x-1+m)}}=x(x+1)...(x+m-1)=x^\overline m$

$$T_0=5;\\2T_n=nT_{n-1}+3\times n!,n>0$$

$a_n=2,b_n=n,c_n=3n!$

$s_n=\frac{a_1a_2...a_n}{b_1b_2...b_n}=\frac{2^n}{n!}$

$S_n=s_1b_1T_0+\sum_{k=1}^ns_kc_k=2\times5+\sum_{k=1}^n\frac{k^k}{k!}\times3k!=3\times 2^{n+1}+4$

$$\sum_{k=0}^{n+1}kH_k=\sum_{k=0}^nkH_k+(n+1)H_{n+1}\\=\sum_{k=0}^n(k+1)H_{k+1}=\sum_{k=0}^n(k+1)(H_k+\frac{1}{k+1})\\=\sum_{k=0}^n(k+1)H_k+\sum_{k=0}^n1$$

$S_n=\sum_{k=0}^n(-1)^{n-k}=\sum_{0\leq n-k\leq n}(-1)^k=[n是偶数]$

$T_n=\sum_{k=0}^n(-1)^{n-k}k=\sum_{0\leq k\leq n}(-1)^k(n-k)=n\sum_{0\leq k\leq n}(-1)^k-\sum_{0\leq k\leq n}(-1)^kk$

$$U_n=\sum_{0\leq k\leq n}(-1)^{n-k}k^2=\sum_{0\leq k\leq n}(-1)^k(n^2-2nk+k^2)\\=n^2\sum_{0\leq k\leq n}(-1)^k-2n\sum_{0\leq k\leq n}(-1)^kk+\sum_{0\leq k\leq n}(-1)^kk^2\\=n^2[n是偶数]-2n\times\frac{n+[n是奇数]}{2}+\frac{(-1)^n(n^+n)}{2}$$

$$\sum_{1\leq j<k\leq n}^{}(a_jb_k-a_kb_j)^2=(\sum_{k=1}^{n}a_k^2)(\sum_{k=1}^{n}b_k^2)-(\sum_{k=1}^{n}a_kb_k)^2$$

$$\sum_{1\leq j<k\leq n}^{}(a_jb_k-a_kb_j)(A_jB_K-A_kB_j)$$

$$(\sum_{k=1}^{n}a_k^2)(\sum_{k=1}^{n}b_k^2)-(\sum_{k=1}^{n}a_kb_k)^2\\=n\sum_{k=1}^{n}(a_kb_k)^2-\sum_{1\leq j<k\leq n}(a_k^2-a^2_j)(b_k^2-b_j^2)-n\sum_{k=1}^n(a_kb_k)^2+\sum_{1\leq j<k\leq n}(a_kb_k-a_jb_j)^2\\=\sum_{1\leq j<k\leq n}((a_kb_k)^2+(a_jb_j)^2-2a_kb_ka_jb_j-((a_kb_k)^2-(a_kb_j)^2-(a_jb_k)^2+(a_jb_j)^2))\\=\sum_{1\leq j<k\leq n}((a_kb_j)^2-2a_kb_ka_jb_j+(a_jb_k)^2)\\=\sum_{1\leq j<k\leq n}(a_jb_k-a_kb_j)^2$$

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

$$\sum\frac{2k+1}{k(k+1)}=\sum(2k+1)(k-1)^\underline{-2}\delta k\\=\sum(2k+1)\Delta(-(x-1)^\underline{-1})=\sum(2k+1)\Delta(-\frac{1}{k})\\=(2k+1)(-\frac{1}{k})+2\sum\frac{1}{k+1}\delta k\\=-2-\frac{1}{k}+2H_k$$

$$\sum\frac{H_k}{(k+1)(k+2)}=\sum H_k\times k^\underline{-2}\delta k\\=-H_kk^\underline{-1}+\sum(k+1)^\underline{-1}k^\underline{-1}\delta k\\=-\frac{H_k}{k+1}+\sum\frac{1}{(k+2)(k+1)}\delta k=-\frac{H_k}{k+1}+\sum k^\underline{-2}\delta k\\=-\frac{H_k}{k+1}-\frac{1}{k+1}=-\frac{H_k+1}{k+1}$$

$\prod_{k\in K}a_k^c=(\prod_{k\in K})^c$

$\prod_{k\in K}a_kb_k=\prod_{k\in K}a_k\times\prod_{k\in K}b_k$

$\prod_{k\in K}a_k=\prod_{p(k)\in K}a_{p(k)}$

$\prod_{j\in J,k\in K}a_{j,k}=\prod_{j\in J}\prod_{k\in K}a_{j,k}$

$\prod_{k\in K}a_k=\prod_ka_k^{[k\in K]}$

$\prod_{k\in K}c=c^{\#k}$

$$P^2=(\prod_{1\leq j,k\leq n}a_ja_k)(\prod_{1\leq j=k\leq n}a_ja_k)\\=(\prod_{1\leq k\leq n}a_k^n)^2(\prod_{1\leq k\leq n}a_k^2)\\=(\prod_{1\leq k\leq n}a_k)^{2n+2}$$

$p=\sqrt{p^2}=(\prod_{1\leq k\leq n}a_k)^{n+1}$

$$\Delta(c^\underline x)=c^{\underline{x+1}}-c^\underline{x}\\=c(c-1)...(c-x+2)-c(c-1)...(c-x+1)\\=c^\underline{x}(c-x-1)=\frac{c^\underline{x+2}}{c-x}$$

$\sum_{k=1}^n\frac{(-2)^\underline{k}}{k}=\sum_{k=1}^{n+1}\frac{(-2)^\underline{k}}{k}\delta k\\=[-(-2)^\underline{k-2}]^{n+1}_1=-(-2)^\underline{n-1}+(-2)^\underline{-1}\\=-1+(-1)(-2)...(-n)\\=(-1)^nn!-1$