# 等比数列

（重定向自等比數列

## 公式

### 等比公式

${\displaystyle q={\frac {a_{n}}{a_{n-1}}}\left(n\geq 2\right)}$

### 通项公式

${\displaystyle a_{2}=a_{1}q}$
${\displaystyle a_{3}=a_{2}q=a_{1}q^{2}}$
${\displaystyle a_{4}=a_{3}q=a_{1}q^{3}}$
${\displaystyle \qquad \quad \vdots }$

${\displaystyle a_{n}=a_{n-1}q=a_{1}q^{n-1}}$

### 求和公式

${\displaystyle a_{1}+a_{2}+a_{3}+\cdots +a_{n}+\cdots }$称为等比数列的和或等比级数，记为${\displaystyle S_{n}}$

${\displaystyle S_{n}=a_{1}+a_{2}+a_{3}+\cdots +a_{n}}$
${\displaystyle \quad \ =a_{1}+a_{1}q+a_{1}q^{2}+\cdots +a_{1}q^{n-1}}$（利用等比数列通项公式）……(1)

${\displaystyle qS_{n}=a_{1}q+a_{1}q^{2}+\cdots +a_{1}q^{n}}$……(2)

(1)式减去(2)式，有：

${\displaystyle (1-q)S_{n}=a_{1}-a_{1}q^{n}}$……(3)

${\displaystyle S_{n}=a_{1}+a_{2}+a_{3}+\cdots +a_{n}}$
${\displaystyle \quad \ =a_{1}+a_{1}q+a_{1}q^{2}+\cdots +a_{1}q^{n-1}}$
${\displaystyle \quad \ =a_{1}+a_{1}+a_{1}+\cdots +a_{1}}$
${\displaystyle \quad \ =na_{1}}$

${\displaystyle S_{n}={\begin{cases}{\frac {a_{1}-a_{1}q^{n}}{1-q}}&q\neq 1\\na_{1}&q=1\end{cases}}}$

${\displaystyle {{S}_{n}}={\frac {a_{1}(1-q^{n})}{1-q}}={\frac {a_{1}q^{n}-a_{1}}{q-1}}}$

### 當-1<q<1時，等比數列無限項之和

${\displaystyle S=\lim _{n\rightarrow \infty }S_{n}=\lim _{n\rightarrow \infty }{\frac {a_{1}-a_{1}q^{n}}{1-q}}={\frac {a_{1}}{1-q}}}$

## 性质

• ${\displaystyle a_{n}=a_{m}q^{n-m}(m,n\in \mathbb {N} ,n>m)}$

• 对于${\displaystyle m,n,s,t\in \mathbb {N} }$，若${\displaystyle m+n=s+t}$，则${\displaystyle a_{m}a_{n}=a_{s}a_{t}}$

${\displaystyle \because m+n=s+t}$
${\displaystyle \therefore a_{m}a_{n}=a_{1}\cdot a_{1}\cdot q^{s+t-2}=a_{1}q^{s-1}\cdot a_{1}q^{t-1}=a_{s}a_{t}}$
• 等比中项：在等比数列中，从第二项起，每一项都是与它等距离的前后两项的等比中项。即等比数列${\displaystyle \left\{a_{n}\right\}}$中有三项${\displaystyle a_{i},a_{j},a_{k}}$，其中${\displaystyle j-i=k-j\geq 1}$，则有${\displaystyle a_{j}^{2}=a_{i}a_{k}}$
• 在原等比数列中，每隔${\displaystyle k}$${\displaystyle (k\in \mathbb {N} )}$取出一项，按原来顺序排列，所得的新数列仍为等比数列。
• ${\displaystyle a_{1},a_{2},a_{3},a_{4},a_{5},a_{6}\cdots }$成等比数列，則${\displaystyle a_{1}a_{2},a_{3}a_{4},a_{5}a_{6}\cdots }$也成等比数列。