求和符号

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求和

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σογμαρω（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：

16 = 1 + 3 + 5 + 7

擴展為數學的一般式：

若有n個數值x₁、x₂、...、x_n，則此n個數值的總和為：

Σ = x₁ + x₂ + ... + x_n

上式的等號右段在數學上常簡潔地寫為：

$\sum _{i=1}^{n}x_{i}$$\sum^{n}_{i=1} x_i$

求和方法

1. 裂項法：利用$a_{k}=b_{k+1}-b_{k}$$a_{k}=b_{k+1}-b_{k}$求出$\sum _{k=m}^{n}a_{k}$$\sum _{k=m}^{n}a_{k}$。
2. 錯位相減法：透過兩個求和式的相減化簡求和數列的求和方法。
3. 倒序求和：對於有對稱中心的函數$f(x)+f(2a-x)=2b$$f(x)+f(2a-x)=2b$首尾求和
4. 逐項求導：可從$\displaystyle \sum _{k=0}^{n}x^{k}={\frac {x^{n+1}-1}{x-1}}$$\displaystyle \sum_{k=0}^n x^k=\frac{x^{n+1}-1}{x-1}$推導出$\displaystyle \sum _{k=0}^{n}k^{m}x^{k}$$\displaystyle \sum_{k=0}^n k^m x^k$
5. 阿貝爾變換：$\sum _{i=1}^{n}a_{i}b_{i}=a_{1}(b_{1}-b_{2})+(a_{1}+a_{2})(b_{2}-b_{3})+\dots +(a_{1}+a_{2}+\dots +a_{n-1})(b_{n-1}-b_{n})+(a_{1}+a_{2}+\dots +a_{n})b_{n}$$\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}$
6. 組合數求和：$\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}}=\sum _{j=1}^{m+1}C_{n}^{j}\Delta ^{j-1}p(1)$$\sum _{k=1}^{n}p(k)={\begin{pmatrix}C_{n}^{1}&C_{n}^{2}&\cdots &C_{n}^{m+1}\end{pmatrix}}{\begin{pmatrix}C_{0}^{0}&0&\cdots &0\\-C_{1}^{0}&C_{1}^{1}&\cdots &0\\\vdots &\vdots &\ddots &\vdots \\(-1)^{m}C_{m}^{0}&(-1)^{m-1}C_{m}^{1}&\cdots &C_{m}^{m}\\\end{pmatrix}}{\begin{pmatrix}p(1)\\p(2)\\\vdots \\p(m+1)\end{pmatrix}}=\sum _{j=1}^{m+1}C_{n}^{j}\Delta ^{j-1}p(1)$
7. 差分算子求逆法：$\displaystyle \sum _{k=1}^{n}p(k)q^{k-1}=f(n)q^{n}-f(0),f(n)={\frac {p(n)}{q-1}}+{\frac {1}{(q-1)^{2}}}\sum _{k=1}^{m}{\frac {(-1)^{k}q^{k-1}}{(q-1)^{k-1}}}\Delta ^{k}(p(n))={\frac {1}{q-1}}\sum _{k=0}^{m}({\frac {-q}{q-1}})^{k}\Delta ^{k}p(n+1)$$\displaystyle \sum _{k=1}^{n}p(k)q^{k-1}=f(n)q^{n}-f(0),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)$

其中$\Delta p(n)=p(n+1)-p(n)$$\Delta p(n)=p(n+1)-p(n)$

常見的總和公式

• 三角形數：$\sum _{i=1}^{n}i={\frac {n(n+1)}{2}}$$\sum^{n}_{i=1} i = \frac{n(n+1)}{2}$
• 正方形數：$\sum _{i=1}^{n}(2i-1)=n^{2}$$\sum_{i=1}^{n} (2i - 1) = n^2$
• 調和級數：$\sum _{n=1}^{k}\,{\frac {1}{n}}\;=\;\ln k+\gamma +\varepsilon _{k}$$\sum_{n=1}^k\,\frac{1}{n} \;=\; \ln k + \gamma + \varepsilon_k$

數列求和公式

• $\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}$$\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}$

• $\sum _{i=0}^{n}x^{i}={\frac {x^{n+1}-1}{x-1}}$$\sum_{i=0}^{n} x^{i} = \frac{x^{n+1}-1}{x-1}$
• 若0 < |x| < 1，則

$\sum _{i=0}^{\infty }x^{i}={\frac {1}{1-x}}$$\sum_{i=0}^{\infty} x^{i} = \frac{1}{1-x}$

• $\displaystyle \sum _{k=1}^{n}[a+(k-1)d]r^{k-1}=[{\frac {a+(n-1)d}{r-1}}-{\frac {d}{(r-1)^{2}}}]r^{n}-[{\frac {a-d}{r-1}}-{\frac {d}{(r-1)^{2}}}]$$\displaystyle \sum _{k=1}^{n}[a+(k-1)d]r^{k-1}=[{\frac {a+(n-1)d}{r-1}}-{\frac {d}{(r-1)^{2}}}]r^{n}-[{\frac {a-d}{r-1}}-{\frac {d}{(r-1)^{2}}}]$

組合數求和公式

• $\sum _{i=0}^{n}{\binom {n}{i}}=2^{n}$$\sum _{{i=0}}^{n}{\binom {n}{i}}=2^{n}$
• $\sum _{i=0}^{n}i^{m}{\binom {n}{i}}=n(n-1)\cdots (n-m+1)2^{n-m}$$\sum _{{i=0}}^{n}i^{m}{\binom {n}{i}}=n(n-1)\cdots (n-m+1)2^{{n-m}}$
• $\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}}$$\sum_{i=m}^n \binom {k_1+i}{k_2} = \binom {k_1+n+1}{k_2+1} - \binom {k_1+m}{k_2+1}$
• $\sum _{i=m}^{n}{\binom {k_{1}+i}{k_{2}+i}}={\binom {k_{1}+n+1}{k_{2}+n}}-{\binom {k_{1}+m}{k_{2}+m-1}}$$\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}$

定積分判斷總和界限

當$f(x)$$f(x)$在[a,b]單調遞增時：

$f(a)+\int _{a}^{b}f(x)dx\leq \sum _{x=a}^{b}f(x)\leq f(b)+\int _{a}^{b}f(x)dx$$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$f(x)$$f(x)$在[a,b]單調遞減時：

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

求和函数

以$\sum _{i=1}^{n}i^{9}$$\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]:= ${\frac {1}{20}}n^{2}(n+1)^{2}\left(n^{2}+n-1\right)\left(2n^{4}+4n^{3}-n^{2}-3n+3\right)$$\frac{1}{20} n^2 (n+1)^2 \left(n^2+n-1\right) \left(2 n^4+4 n^3-n^2-3 n+3\right)$