求和符号

求和符号（Σ，sigma），是欧拉于1755年首先使用的。这个符号是源于希腊文（增加）的字头，Σ正是σ的大写。求和的结果是給定的數值相加後的總值，又稱加總。

舉例而言，若有4個數值：1、3、5、7，則這4個數值的總和為：

${\displaystyle 16=1+3+5+7}$

擴展為數學的一般式：若有${\displaystyle n}$個數值${\displaystyle x_{1},x_{2},\cdots ,x_{n}}$，則此${\displaystyle n}$個數值的總和為：

${\displaystyle \Sigma =x_{1}+x_{2}+\cdots +x_{n}}$

上式的等號右段在數學上常簡潔地寫為：

${\displaystyle \sum _{k=1}^{n}x_{k}}$

上式意思为${\displaystyle x_{k}}$项(${\displaystyle k=1}$，即${\displaystyle x_{k}=x_{1}}$)到${\displaystyle x_{n}}$项的求和。

求和方法

1. 裂項法：利用${\displaystyle a_{k}=b_{k+1}-b_{k}}$求出${\displaystyle \sum _{k=m}^{n}a_{k}}$。
2. 錯位相減法：透過兩個求和式的相減化簡求和數列的求和方法。
3. 倒序求和：對於有對稱中心的函數${\displaystyle f(x)+f(2a-x)=2b}$首尾求和[1][2]
4. 逐項求導：可從${\displaystyle \displaystyle \sum _{k=0}^{n}x^{k}={\frac {x^{n+1}-1}{x-1}}}$推導出${\displaystyle \displaystyle \sum _{k=0}^{n}k^{m}x^{k}}$[3]
5. 阿貝爾變換：

${\displaystyle \sum _{i=1}^{n}a_{i}b_{i}=a_{1}(b_{1}-b_{2})+(a_{1}+a_{2})(b_{2}-b_{3})+\dots +(a_{1}+a_{2}+\dots +a_{n-1})(b_{n-1}-b_{n})+(a_{1}+a_{2}+\dots +a_{n})b_{n}}$

含多項式求和公式

以下設p為多項式，${\displaystyle \deg p(k)=m,\Delta p(k)=p(k+1)-p(k)}$

${\displaystyle \sum p(k)}$

${\displaystyle \sum p(k)}$是對一個多項式求和，自然數方冪和、等幂求和、等差數列求和都屬于對多項式求和。

• 帕斯卡矩陣形式

  ${\displaystyle \sum _{k=1}^{n}p(k)={\begin{pmatrix}C_{n}^{1}&C_{n}^{2}&\cdots &C_{n}^{m+1}\end{pmatrix}}{\begin{pmatrix}C_{0}^{0}&0&\cdots &0\\-C_{1}^{0}&C_{1}^{1}&\cdots &0\\\vdots &\vdots &\ddots &\vdots \\(-1)^{m}C_{m}^{0}&(-1)^{m-1}C_{m}^{1}&\cdots &C_{m}^{m}\\\end{pmatrix}}{\begin{pmatrix}p(1)\\p(2)\\\vdots \\p(m+1)\end{pmatrix}}}$[4]

• 差分變換形式

  ${\displaystyle p(k)=\sum _{j=1}^{m+1}C_{k-1}^{j-1}\Delta ^{j-1}p(1)}$

  ${\displaystyle \sum _{k=1}^{n}p(k)=\sum _{j=1}^{m+1}C_{n}^{j}\Delta ^{j-1}p(1)}$[5]

${\displaystyle \sum p(k)}$的例子

• ${\displaystyle \sum _{k=1}^{n}k^{0}=\sum _{k=1}^{n}1=n}$
• 三角形數：${\displaystyle \sum _{k=1}^{n}k={\frac {n(n+1)}{2}}}$
• 等差級數：${\displaystyle \sum _{i=0}^{n-1}(a_{1}+id)={\frac {n(a_{1}+a_{n})}{2}}={\frac {n[2a_{1}+(n-1)d]}{2}}=a_{1}C_{n}^{1}+dC_{n}^{2}}$
• 連續正整數平方和：${\displaystyle \sum _{k=1}^{n}k^{2}={\frac {n(n+1)(2n+1)}{6}}}$
• 連續正整數立方和：${\displaystyle \sum _{k=1}^{n}k^{3}=\left[{\frac {n(n+1)}{2}}\right]^{2}}$
• 正方形數：${\displaystyle \sum _{k=1}^{n}(2k-1)=n^{2}}$

${\displaystyle \sum u_{k}v_{k}x^{k}}$

當${\displaystyle u_{k}=p(k)}$為多項式，${\displaystyle \sum _{l=0}^{\infty }v_{l}x^{l}}$易求高階導數時，${\displaystyle \sum _{k=0}^{\infty }u_{k}v_{k}x^{k}}$有封閉型和式

${\displaystyle \sum _{k=0}^{\infty }u_{k}v_{k}x^{k}=\sum _{k=0}^{\infty }{\frac {\Delta ^{k}u_{0}x^{k}}{k!}}{\frac {d^{k}}{dx^{k}}}(\sum _{l=0}^{\infty }v_{l}x^{l})}$[6]

${\displaystyle \sum p(k)q^{k}}$

• ${\displaystyle u_{k}=p(k),v_{k}=1,x=q,\sum u_{k}v_{k}x^{k}=\sum p(k)q^{k}}$

  有限和${\displaystyle \displaystyle \sum _{k=1}^{n}p(k)q^{k-1}}$有封閉型和式

  當p為常數時，是對等比數列求和，當p為一次多項式時，是對差比數列求和。

  ${\displaystyle \displaystyle \sum _{k=1}^{n}p(k)q^{k-1}=f(n)q^{n}-f(0)}$

  ${\displaystyle f(n)={\frac {p(n)}{q-1}}+{\frac {1}{(q-1)^{2}}}\sum _{k=1}^{m}{\frac {(-1)^{k}q^{k-1}}{(q-1)^{k-1}}}\Delta ^{k}(p(n))={\frac {1}{q-1}}\sum _{k=0}^{m}({\frac {-q}{q-1}})^{k}\Delta ^{k}p(n+1)}$[7]

${\displaystyle \sum p(k)q^{k}}$的例子

• 等比級數：${\displaystyle \sum _{i=0}^{n-1}a_{1}x^{i}={\frac {a_{1}(x^{n}-1)}{x-1}}}$，若${\displaystyle 0<|x|<1}$，則${\displaystyle \sum _{i=0}^{\infty }a_{1}x^{i}={\frac {a_{1}}{1-x}}}$
• 差比級數：${\displaystyle \displaystyle \sum _{k=1}^{n}[a+(k-1)d]r^{k-1}=\left[{\frac {a+(n-1)d}{r-1}}-{\frac {d}{(r-1)^{2}}}\right]r^{n}-\left[{\frac {a-d}{r-1}}-{\frac {d}{(r-1)^{2}}}\right]}$

${\displaystyle \sum {\frac {p(k)}{k!}}x^{k}}$

• ${\displaystyle u_{k}=p(k),v_{k}={\frac {1}{k!}},\sum u_{k}v_{k}x^{k}=\sum {\frac {p(k)}{k!}}x^{k}}$

  ${\displaystyle \sum _{n=0}^{\infty }{\frac {p(n)}{n!}}x^{n}=e^{x}\sum _{k=0}^{m}{\frac {\Delta ^{k}p(0)}{k!}}x^{k}}$[8]

${\displaystyle \sum H_{k}p(k)}$

${\displaystyle \sum _{k=1}^{n}H_{k}p(k)=(\sum _{j=0}^{m}C_{n+1}^{j+1}\Delta ^{j}p(0))H_{n}-\sum _{j=0}^{m}{\frac {C_{n}^{j+1}}{j+1}}\Delta ^{j}p(0)}$，其中${\displaystyle H_{n}}$為調和數或調和級數

組合數求和公式

一阶求和公式

• ${\displaystyle \sum _{r=0}^{n}{\binom {n}{r}}=2^{n}}$
• ${\displaystyle \sum _{r=0}^{n-k}{\frac {(-1)^{r}(n+1)}{k+r+1}}{\binom {n-k}{r}}={\binom {n}{k}}^{-1}}$
• ${\displaystyle \sum _{r=0}^{n}{\binom {dn}{dr}}={\frac {1}{d}}\sum _{r=1}^{d}(1+e^{\frac {2\pi ri}{d}})^{dn}}$[参 1]

• ${\displaystyle F_{n}=\sum _{i=0}^{\infty }{\binom {n-i}{i}}}$[参 2]

${\displaystyle F_{n-1}+F_{n}=\sum _{i=0}^{\infty }{\binom {n-1-i}{i}}+\sum _{i=0}^{\infty }{\binom {n-i}{i}}=1+\sum _{i=1}^{\infty }{\binom {n-i}{i-1}}+\sum _{i=1}^{\infty }{\binom {n-i}{i}}=1+\sum _{i=1}^{\infty }{\binom {n+1-i}{i}}=\sum _{i=0}^{\infty }{\binom {n+1-i}{i}}=F_{n+1}}$

• ${\displaystyle \sum _{i=m}^{n}{\binom {i}{a}}={\binom {n+1}{a+1}}-{\binom {m}{a+1}}}$

${\displaystyle {\binom {m}{a+1}}+{\binom {m}{a}}+{\binom {m+1}{a}}...+{\binom {n}{a}}={\binom {n+1}{a+1}}}$

${\displaystyle \sum _{i=m}^{n}{\binom {k_{1}+i}{k_{2}}}={\binom {k_{1}+n+1}{k_{2}+1}}-{\binom {k_{1}+m}{k_{2}+1}}}$

${\displaystyle \sum _{i=m}^{n}{\binom {k_{1}+i}{k_{2}+i}}={\binom {k_{1}+n+1}{k_{2}+n}}-{\binom {k_{1}+m}{k_{2}+m-1}}}$

二阶求和公式

• ${\displaystyle \sum _{r=0}^{n}{\binom {n}{r}}^{2}={\binom {2n}{n}}}$
• ${\displaystyle \sum _{i=0}^{n}{\binom {r_{1}+n-1-i}{r_{1}-1}}{\binom {r_{2}+i-1}{r_{2}-1}}={\binom {r_{1}+r_{2}+n-1}{r_{1}+r_{2}-1}}}$[参 3]

${\displaystyle (1-x)^{-r_{1}}(1-x)^{-r_{2}}=(1-x)^{-r_{1}-r_{2}}}$

${\displaystyle (1-x)^{-r_{1}}(1-x)^{-r_{2}}=(\sum _{n=0}^{\infty }{\binom {r_{1}+n-1}{r_{1}-1}}x^{n})(\sum _{n=0}^{\infty }{\binom {r_{2}+n-1}{r_{2}-1}}x^{n})=\sum _{n=0}^{\infty }(\sum _{i=0}^{n}{\binom {r_{1}+n-1-i}{r_{1}-1}}{\binom {r_{2}+i-1}{r_{2}-1}})x^{n}}$

${\displaystyle (1-x)^{-r_{1}-r_{2}}=\sum _{n=0}^{\infty }{\binom {r_{1}+r_{2}+n-1}{r_{1}+r_{2}-1}}x^{n}}$

• ${\displaystyle \sum _{i=0}^{k}{\binom {n}{i}}{\binom {m}{k-i}}={\binom {n+m}{k}}}$

范德蒙恒等式與超幾何函數有關係：

${\displaystyle \sum _{i=0}^{k}{\binom {n}{i}}{\binom {m}{k-i}}={\frac {m!}{k!(m-k)!}}{}_{2}F_{1}(-n,-k;m-k+1;1)={\binom {n+m}{k}}}$

三阶求和公式

• ${\displaystyle {\binom {n+k}{k}}^{2}=\sum _{j=0}^{k}{\binom {k}{j}}^{2}{\binom {n+2k-j}{2k}}}$

范德蒙恒等式與廣義超幾何函數有關係：

${\displaystyle \sum _{j=0}^{k}{\binom {k}{j}}^{2}{\binom {n+2k-j}{2k}}={\frac {(n+2k)!}{(2k)!n!}}{}_{3}F_{2}(-k,-k,-n;1,-n-2k;1)={\binom {n+k}{k}}^{2}}$

定積分判斷總和界限

當${\displaystyle f(x)}$在[a,b]單調遞增時：

${\displaystyle f(a)+\int _{a}^{b}f(x)dx\leq \sum _{x=a}^{b}f(x)\leq f(b)+\int _{a}^{b}f(x)dx}$

當${\displaystyle f(x)}$在[a,b]單調遞減時：

${\displaystyle f(b)+\int _{a}^{b}f(x)dx\leq \sum _{x=a}^{b}f(x)\leq f(a)+\int _{a}^{b}f(x)dx}$[9]

求和函数

以${\displaystyle \sum _{i=1}^{n}i^{9}}$为例：

• Matlab

syms k n;symsum(k^9,k,1,n)

• Mathematica

 In[1]:= Sum[i^9, {i, 1, n}]
 Out[1]:= ${\displaystyle {\frac {1}{20}}n^{2}(n+1)^{2}\left(n^{2}+n-1\right)\left(2n^{4}+4n^{3}-n^{2}-3n+3\right)}$

参考资料

1. 赵丽棉 黄基廷. . 高等数学研究. 2010, (4).
2. 徐更生 何廷模. . 中学教研. 1991, (10).
3. 伍启期. . 佛山科学技术学院学报(自然科学版). 1996, (4).

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