# 等比数列

${\displaystyle 3,6,12,24,48,96,...}$

## 性质

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

${\displaystyle \{a\,,\,\,ar\,,\,\,ar^{2}\,,\,\cdots \,,\,\,ar^{n-1}\}}$

${\displaystyle r={\frac {a_{n+1}}{a_{n}}}}$

${\displaystyle r={\sqrt[{m-n}]{\frac {a_{m}}{a_{n}}}}}$

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

{\displaystyle {\begin{aligned}a_{n-1}\times a_{n+1}&=ar^{n-2}\times ar^{n}\\&=a^{2}\times r^{2n-2}\\&=(ar^{n-1})^{2}\\&={a_{n}}^{2}\\\end{aligned}}}

${\displaystyle a_{n}=\pm {\sqrt {a_{n-1}\cdot a_{n+1}}}}$

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

{\displaystyle {\begin{aligned}a_{m}\cdot a_{n}&=ar^{m-1}\cdot ar^{n-1}\\&=a^{2}r^{m+n-2}\\&=a^{2}r^{p+q-2}\\&=ar^{p-1}\cdot ar^{q-1}\\&=a_{p}\cdot a_{q}\\\end{aligned}}}

${\displaystyle a_{n-k}\cdot a_{n+k}={a_{n}}^{2}}$
${\displaystyle a_{n}=\pm {\sqrt {a_{n-k}\cdot a_{n+k}}}}$

• ${\displaystyle \{b\cdot a_{n}\}}$ 是一个等比数列。
• ${\displaystyle \{{\frac {b}{a_{n}}}\}}$ 是一个等比数列。
• ${\displaystyle \{\log _{b}(a_{n})\}}$ 是一个等差数列

${\displaystyle a_{n}=pq^{n}}$

## 等比数列和

${\displaystyle S_{n}={\frac {a(1-r^{n})}{1-r}}}$

${\displaystyle S_{n}=a+ar+ar^{2}+\cdots +ar^{n-1}}$ ……(1)

${\displaystyle rS_{n}=ar+ar^{2}+\cdots +ar^{n}}$ ……(2)

(1)式减去(2)式，有：

${\displaystyle (1-r)S_{n}=a-ar^{n}}$

${\displaystyle r\neq 1}$时，整理后得证。

${\displaystyle r=1}$时，可以发现：

{\displaystyle {\begin{aligned}S_{n}&=a+ar+ar^{2}+\cdots +ar^{n-1}\\&={\begin{matrix}\underbrace {a+a+a+\cdots +a} \\n\end{matrix}}\\&=n\times a\\&=an\\\end{aligned}}}

${\displaystyle S_{n}={\begin{cases}{\frac {a(1-r^{n})}{1-r}}&r\neq 1\\an&r=1\end{cases}}}$

${\displaystyle -1时，注意到

${\displaystyle \lim _{n\rightarrow \infty }r^{n}=0}$

${\displaystyle S_{\infty }={\frac {a}{1-r}}}$

## 等比数列积

${\displaystyle P_{n}=a^{n}\cdot r^{\frac {n(n-1)}{2}}}$

{\displaystyle {\begin{aligned}P_{n}&=a\cdot ar\cdot ar^{2}\cdot \cdots \cdot ar^{n-1}\\&=a^{n}\cdot r^{1+2+\cdots +(n-1)}\\&=a^{n}\cdot r^{\frac {n(n-1)}{2}}\\\end{aligned}}}