# 等差-等比数列

## 通项公式

${\displaystyle [a+(n-1)d]r^{n-1}}$

${\displaystyle [a+(n)(1)d]}$

## 等差-等比数列的求和公式

${\displaystyle \sum _{k=1}^{n}\left[a+(k-1)d\right]r^{k-1}=a+(a+d)r+(a+2d)r^{2}+\cdots +[a+(n-1)d]r^{n-1}}$

${\displaystyle S_{n}=\sum _{k=1}^{n}\left[a+(k-1)d\right]r^{k-1}={\frac {a}{1-r}}-{\frac {[a+(n-1)d]r^{n}}{1-r}}+{\frac {dr(1-r^{n-1})}{(1-r)^{2}}}.}$

### 错位相减法

${\displaystyle S_{n}=a+(a+d)r+(a+2d)r^{2}+\cdots +[a+(n-1)d]r^{n-1}}$

Sn乘以r

${\displaystyle rS_{n}=ar+(a+d)r^{2}+(a+2d)r^{3}+\cdots +[a+(n-1)d]r^{n}}$

Sn减去rSn

{\displaystyle {\begin{aligned}S_{n}(1-r)&=&\left\{a+(a+d)r+(a+2d)r^{2}+\cdots +[a+(n-1)d]r^{n-1}\right\}\\&&-\left\{ar+(a+d)r^{2}+(a+2d)r^{3}+\cdots +[a+(n-1)d]r^{n}\right\}\\&=&a+\left[dr+dr^{2}+\cdots +dr^{n-1}\right]-[a+(n-1)d]r^{n}\\&=&a+\left[{\frac {dr(1-r^{n-1})}{1-r}}\right]-[a+(n-1)d]r^{n}\end{aligned}}}

### 逐项求导

${\displaystyle \displaystyle \sum _{k=1}^{n}r^{k-1}={\frac {r^{n}-1}{r-1}}}$

${\displaystyle \displaystyle \sum _{k=1}^{n}(k-1)r^{k-2}={\frac {nr^{n-1}}{r-1}}-{\frac {r^{n}-1}{(r-1)^{2}}}}$

${\displaystyle \displaystyle \sum _{k=1}^{n}[a+(k-1)d]r^{k-1}=a{\frac {r^{n}-1}{r-1}}+dr[{\frac {nr^{n-1}}{r-1}}-{\frac {r^{n}-1}{(r-1)^{2}}}]}$

### 裂项法

${\displaystyle [a+(k-1)d]r^{k-1}=(sk+t)r^{k}-[s(k-1)+t]r^{k-1}}$

${\displaystyle \displaystyle \sum _{k=1}^{n}[a+(k-1)d]r^{k-1}=(sn+t)r^{n}-t}$

${\displaystyle dk+a-d=s(r-1)k+(r-1)t+s}$

${\displaystyle \displaystyle s={\frac {d}{r-1}},t={\frac {a-d}{r-1}}-{\frac {d}{(r-1)^{2}}}}$

${\displaystyle \displaystyle \sum _{k=1}^{n}[a+(k-1)d]r^{k-1}=[{\frac {d}{r-1}}n+{\frac {a-d}{r-1}}-{\frac {d}{(r-1)^{2}}}]r^{n}-[{\frac {a-d}{r-1}}-{\frac {d}{(r-1)^{2}}}]}$

### 差分算子公式

${\displaystyle \displaystyle \sum _{k=1}^{n}p(k)q^{k-1}=f(n)q^{n}-f(0),f(n)={\frac {p(n)}{q-1}}+{\frac {1}{(q-1)^{2}}}\sum _{k=1}^{m}{\frac {(-1)^{k}q^{k-1}}{(q-1)^{k-1}}}\Delta ^{k}(p(n))={\frac {1}{q-1}}\sum _{k=0}^{m}({\frac {-q}{q-1}})^{k}\Delta ^{k}p(n+1)}$[5]

${\displaystyle \displaystyle f(n)={\frac {a+(n-1)d}{r-1}}-{\frac {d}{(r-1)^{2}}}}$

${\displaystyle \displaystyle \sum _{k=1}^{n}[a+(k-1)d]r^{k-1}=[{\frac {a+(n-1)d}{r-1}}-{\frac {d}{(r-1)^{2}}}]r^{n}-[{\frac {a-d}{r-1}}-{\frac {d}{(r-1)^{2}}}]}$

## 无穷级数

${\displaystyle \lim _{n\to \infty }S_{n}={\frac {a}{1-r}}+{\frac {dr}{(1-r)^{2}}}}$

## 参考文献

1. K.F. Riley, M.P. Hobson, S.J. Bence. Mathematical methods for physics and engineering 3rd. Cambridge University Press. 2010: 118. ISBN 978-0-521-86153-3.
2. ^ 江凤莲. 利用“错位相减法”解数列问题. 龙岩师专学报. 2001, (S1).
3. ^ 李曰玮 刘瑞楼. 一类特殊多项式的求和问题. 高等数学研究. 2012, (1).
4. ^ 郑良. 差比型数列前n项和的求解方法——裂项法. 中学生数学. 2012, (3).
5. ^ 黄嘉威. 方幂和及其推广和式. 数学学习与研究. 2016, (7).