# 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

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

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

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

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

$C^n_k=C^{n-1}_k+C^{n-1}_{k-1}$帕斯卡法則

$\frac{1}{C^n_k}=\frac{n+1}{n+2}\left(\frac{1}{C^{n+1}_k}+\frac{1}{C^{n+1}_{k+1}}\right)$[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

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

## 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.