数列

Sequence
一系列按照一定顺序排列的数，无穷数列的项之和构成级数
按照某一法则，对每个 $n \in N_{+}$ 对应着一个确定的实数 $x_{n}$
按照下表 $n$ 从小到大排列得到的一个序列:

\begin{array}{r} x_{1}, x_{2}, \dots, x_{n}, \dots \end{array}

就叫做数列，简记为 ${x_{n}}$

\begin{array}{r} \sum_{i = 1}^{n} i = \frac{n (n + 1)}{2} \end{array}

\begin{array}{r} \sum_{i = 1}^{n} i^{2} = \frac{n (n + 1) (2 n + 1)}{6} \end{array}

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

\begin{array}{r} S_{n} = \frac{n (a_{1} + a_{n})}{2} = n a_{1} + \frac{n (n - 1)}{2} d \end{array}

等比级数，又称几何级数

$\frac{a_{n + 1}}{a_{n}} = q$
$a_{n} = a_{1} q^{n - 1}$

S_{n} = {\begin{cases} n a_{1} & q = 1 \\ \frac{a_{1} (1 - q^{n})}{1 - q} = \frac{a_{1} - a_{n} q}{1 - q} & q \neq 1 \end{cases}

错位相减法
$S_{n} = a_{1}, a_{1} q, \dots, a_{1} q^{n - 1}$
$q S_{n} = a_{1} q, a_{1} q^{2}, \dots, a_{1} q^{n}$