# 等差-等比数列

## 通项公式

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

$[a+(n-1)d]$

## 等差-等比数列的求和公式

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

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

### 导出

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

Sn乘以r

$r S_n = ar+(a+d)r^2+(a+2d)r^3+\cdots +(a+(n-1))r^n$

Sn减去rSn

\begin{align} S_n(1-r) &=&\left[a+(a+d)r+(a+2d)r^2+\cdots +[a+(n-1)]r^{n-1}\right] \\ & &- \left[ar+(a+d)r^2+(a+2d)r^3+\cdots + [ a+(n-1) ] r^n\right] \\ & = & a+ \left[ rd + r^2d + \cdots \right] - [ a+d(n-1) ] r^n \\ & = & a + \left[ \frac{rd(1-r^{n-1})}{1-r}\right]-[a+(n-1)d]r^n\end{align}

## 无穷级数

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

## 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. ^ 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.
3. ^ 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.