# 等差数列

3, 5, 7, 9, 11, 13, ...

## 性質

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

${\displaystyle \{a\,,\,\,a+d\,,\,\,a+2d\,,\,\cdots \,,\,\,a+(n-1)d\}}$

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

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

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

{\displaystyle {\begin{aligned}a_{n-1}+a_{n+1}&=[a+(n-2)d]+(a+nd)\\&=2a+(2n-2)d\\&=2[a+(n-1)d]\\&=2a_{n}\\\end{aligned}}}

${\displaystyle a_{n}={\frac {a_{n-1}+a_{n+1}}{2}}}$

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

{\displaystyle {\begin{aligned}a_{m}+a_{n}&=[a+(m-1)d]+[a+(n-1)d]\\&=2a+(m+n-2)d\\&=2a+(p+q-2)d\\&=[a+(p-1)d]+[a+(q-1)d]\\&=a_{p}+a_{q}\\\end{aligned}}}

${\displaystyle a_{n-k}+a_{n+k}=2a_{n}}$
${\displaystyle a_{n}={\frac {a_{n-k}+a_{n+k}}{2}}}$

• ${\displaystyle \{b+a_{n}\}}$ 是一個等差數列。
• ${\displaystyle \{b\cdot a_{n}\}}$ 是一個等差數列。
• ${\displaystyle \{b^{a_{n}}\}}$ 是一個等比數列
• ${\displaystyle \{{\frac {b}{a_{n}}}\}}$ 是一個等諧數列

${\displaystyle a_{n}=p+qn}$

## 等差數列和

{\displaystyle {\begin{aligned}S_{n}&={\frac {n}{2}}\,(a+a_{n})\\&={\frac {n}{2}}[2a+(n-1)d]\\&=an+d\cdot {\frac {n(n-1)}{2}}\end{aligned}}}

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

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

${\displaystyle S_{n}=pn+qn^{2}}$

## 等差数列积

${\displaystyle P_{n}=d^{n}\cdot {\frac {\Gamma ({\frac {a}{d}}+n)}{\Gamma ({\frac {a}{d}})}}}$

{\displaystyle {\begin{aligned}P_{n}&=a\cdot (a+d)\cdot (a+2d)\cdot \cdots \cdot [a+(n-1)d]\\&=d^{n}\cdot \left({\frac {a}{d}}\right)\cdot \left({\frac {a}{d}}+1\right)\cdot \left({\frac {a}{d}}+2\right)\cdot \cdots \cdot \left[{\frac {a}{d}}+(n-1)\right]\\&=d^{n}\cdot {\left({\frac {a}{d}}\right)}^{\overline {n}}\\&=d^{n}\cdot {\frac {\Gamma ({\frac {a}{d}}+n)}{\Gamma ({\frac {a}{d}})}}\\\end{aligned}}}

{\displaystyle {\begin{aligned}P_{4}&=2^{4}\cdot {\frac {\Gamma ({\frac {1}{2}}+4)}{\Gamma ({\frac {1}{2}})}}\\&=16\cdot {\frac {11.6317\dots }{1.77245\dots }}\\&=105\end{aligned}}}

## 注释

1. ^ 也有人使用arithmetic progression，簡稱A.P.