数列

等差数列

$a_n=a_1+(n-1)d \\S_n=\frac{n(a_1+a_n)}2=na_1+\frac{n(n-1)}2d$

等比数列

$a_n=a_1q^{n-1}$ $S_n=\begin{cases}\frac{a_1(1-q^n )}{1-q },(q\ne1)\\na_1,(q=1)\end{cases}\\ \lim_{x \to \infty} S_n=\frac{a_1}{1-q}$

通项

不动点

$a_{n+1}=pa_n+q \Rightarrow x_0=px_0+q \Rightarrow a_{n+1}-x_0=p(a_n-x_0)$

待定系数

$a_{n+2}=pa_{n+1}+qa_{n}\\ \Rightarrow a_{n+2}+\lambda a_{n+1}=(p+\lambda)(a_{n+1}+\lambda a_n)$

求和

公式

$\sum_{k=1}^nk^2=\frac{n(n+1)(2n+1)}6\\ \sum_{k=1}^nk^3=\left [ \frac{n(n+1)}{2}\right ]^2$

裂项

$\frac1{n^2}>\frac1{n(n+1)}=\frac{(n+1)-n}{n(n+1)}=\frac1n-\frac1{n+1}$ $\frac{1}{\sqrt {n + 1} + \sqrt n } = \sqrt {n + 1} - \sqrt n$

错位相减

$a_n=a_1+(n-1)d\quad\quad b_n=b_1q^{n-1}$ $\begin{cases}T_{n}=a_{1} b_{1}+a_{2} b_{2}+a_{3} b_{3}+\cdots+a_{n-1} b_{n-1}+a_{n} b_{n}\\ q T_{n}=a_{1} b_{2}+a_{2} b_{3}+a_{3} b_{4}+\cdots+a_{n-1} b_{n}+q a_{n} b_{n}\end{cases}$ $\begin{aligned} T_{n}-q T_{n} &=a_{1} b_{1}+d\left(b_{2}+b_{2}+\cdots+b_{n}\right)-q a_{n} b_{n} \\ &=a_{1} b_{1}+d\left(\frac{b_{1}\left(1-q^{n}\right)}{1-q}-b_{1}\right)-q a_{n} b_{n} \\ &=a_{1} b_{1}+d\left(S_{n}-b_{1}\right)-q a_{n} b_{n} \\ &=\left(a_{1}-d\right) b_{1}+d S_{n}-q a_{n} b_{n} \end{aligned}$ $T_{n}=\frac{\left(a_{1}-d\right) b_{1}+d S_{n}-q a_{n} b_{n}}{1-q}$