# 裂項和

$\sum_{i=1}^n(a_i-a_{i+1}) = (a_1-a_2)+(a_2-a_3)+\ldots+(a_n-a_{n+1})=a_1-a_{n+1}.$

$\prod_{i=1}^n \frac{a_i}{a_{i+1}} = \frac{a_1}{a_{n+1}}$

## 可以用來裂项求和的數學式

$\frac{1}{a(a+b)}=\frac{1}{b}(\frac{1}{a}-\frac{1}{a+b})$

$a^k=\frac{1}{a-1}(a^{k+1}-a^k)$

$coskx=\frac{1}{2sin\frac{x}{2}}[sin(k+\frac{1}{2})x-sin(k-\frac{1}{2})x]$

$sinkx=\frac{1}{2sin\frac{x}{2}}[cos(k-\frac{1}{2})x-cos(k-\frac{1}{2})x]$三角恒等式[1]

$C_n^k=C_{n-1}^k+C_{n-1}^{k-1}$帕斯卡法則

$\frac{1}{C_n^k}=\frac{n+1}{n+2}(\frac{1}{C_{n+1}^k}+\frac{1}{C_{n+1}^{k+1}})$[2]

## 求和类型

### 一般求和

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

### 交错求和

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

## 誤用

$0 = \sum_{n=1}^\infty 0 = \sum_{n=1}^\infty (1-1) = 1 + \sum_{n=1}^\infty (-1 + 1) = 1\,$

$\sum_{n=1}^N \frac{1}{n(n+1)} = \sum_{n=1}^N \frac{1}{n} - \frac{1}{(n+1)}\,$
$= \left(1 - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \cdots + \left(\frac{1}{N} - \frac{1}{N+1}\right)\,$
$= 1 + \left(- \frac{1}{2} + \frac{1}{2}\right) + \left( - \frac{1}{3} + \frac{1}{3}\right) + \cdots + \left(-\frac{1}{N} + \frac{1}{N}\right) - \frac{1}{N+1} \,$
$= 1 - \frac{1}{N+1}\to 1\ \mathrm{as}\ N\to\infty.\,$

## 例子：三角函數

$\sum_{n=1}^N \sin\left(n\right) = \sum_{n=1}^N \frac{1}{2} \csc\left(\frac{1}{2}\right) \left(2\sin\left(\frac{1}{2}\right)\sin\left(n\right)\right)$
$=\frac{1}{2} \csc\left(\frac{1}{2}\right) \sum_{n=1}^N \left(\cos\left(\frac{2n-1}{2}\right)-\cos\left(\frac{2n+1}{2}\right)\right)$
$=\frac{1}{2} \csc\left(\frac{1}{2}\right) \left(\cos\left(\frac{1}{2}\right)-\cos\left(\frac{2N+1}{2}\right)\right).$

## 參考資料

1. ^
2. ^ 及万会 张来萍 杨春艳. 封闭形和式初步.