# 等差數列

## 通項公式

 ${\displaystyle \ a_{n}=a_{1}+(n-1)d}$

${\displaystyle \ a_{n}=a_{m}+(n-m)d}$

${\displaystyle \ d={\frac {a_{n}-a_{m}}{n-m}}}$

## 等差中項

${\displaystyle a_{n-1}+a_{n+1}\neq 2a_{n}}$

${\displaystyle a_{1}+(n-2)d+a_{1}+(n)d\neq 2[a_{1}+(n-1)d]}$
${\displaystyle 2a_{1}+2nd-2d\neq 2a_{1}+2nd-2d}$(矛盾)
${\displaystyle \ a_{n-1}+a_{n+1}=2a_{n}}$

## 等差數列的和

### 公式

${\displaystyle S_{n}=a_{1}+a_{2}+\dots +a_{n}=\sum _{i=0}^{n-1}(a_{1}+id)={\frac {n(a_{1}+a_{n})}{2}}={\frac {n[2a_{1}+(n-1)d]}{2}}.}$

### 證明

${\displaystyle S_{n}=a_{1}+(a_{1}+d)+(a_{1}+2d)+\dots +[a_{1}+(n-2)d]+[a_{1}+(n-1)d]}$
${\displaystyle S_{n}=[a_{n}-(n-1)d]+[a_{n}-(n-2)d]+\dots +(a_{n}-2d)+(a_{n}-d)+a_{n}}$

（往左滑）

${\displaystyle \ 2S_{n}=n(a_{1}+a_{n})}$

 ${\displaystyle S_{n}={\frac {n(a_{1}+a_{n})}{2}}={\frac {n[2a_{1}+(n-1)d]}{2}}}$

### 幾何方法

${\displaystyle 1+2+3+\dots +10={\frac {(1+10)+(2+9)+(3+8)+\dots +(10+1)}{2}}={\frac {(1+10)\times 10}{2}}=55.}$

${\displaystyle S_{n}=a_{1}+a_{2}+\dots +a_{n}}$
${\displaystyle S_{n}=a_{1}+(a_{1}+d)+(a_{1}+2d)+\dots \dots +[a_{1}+(n-2)d]+[a_{1}+(n-1)d]}$
${\displaystyle S_{n}={\frac {\left\{[a_{1}+[a_{1}+(n-1)d]\right\}\times n}{2}}={\frac {[2a_{1}+(n-1)d]\times n}{2}}}$

## 等差數列的積

${\displaystyle a_{1}a_{2}\cdots a_{n}=d^{n}{\left({\frac {a_{1}}{d}}\right)}^{\overline {n}}}$

## 其他性質

${\displaystyle a_{m}+a_{n}=a_{p}+a_{q}}$

     ${\displaystyle S_{m+n}=a(m+n)+{\frac {(m+n)(m+n-1)d}{2}}}$
${\displaystyle {\frac {S_{m+n}(m-n)}{m+n}}=a(m-n)+{\frac {(m^{2}-n^{2}-m+n)d}{2}}}$
${\displaystyle {\frac {S_{m+n}(m-n)}{m+n}}=am+{\frac {m(m-1)d}{2}}-an-{\frac {n(n-1)d}{2}}=S_{m}-S_{n}}$
${\displaystyle S_{m+n}={\frac {(S_{m}-S_{n})(m+n)}{m-n}}}$


## 參考文獻

• Sigler, Laurence E. (trans.). Fibonacci's Liber Abaci. Springer-Verlag. 2002: 259–260. ISBN 978-0-387-95419-6.