等差-等比数列

等差-等比数列有如下通项公式：^[1]

${\displaystyle [a+(n-1)d]r^{n-1}}$ 

其中${\displaystyle r}$ 是公比，而${\displaystyle r^{n-1}}$ 的系数：

${\displaystyle [a+(n-1)d]}$ 

则是等差数列的项，其首项为${\displaystyle a}$ ，公差${\displaystyle d}$ 。

等差-等比级数有如下形式；

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

其前n项之和为；

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

错位相减法

由此级数开始：^[1][2]

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

将S_n乘以r，

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

S_n减去rS_n，

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

在中间的项中使用等比数列的求和公式。最后左右两边同除以(1 − r)，得到最终结果。

逐项求导

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

对等比数列和两边求导：^[3]

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

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

裂项法

待定系数s,t使得等差-等比数列可以裂项：^[4]

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

用裂项法可以求出数列和：

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

求出待定系数s,t关于a,d,r的表达式：

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

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

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

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]

其中${\displaystyle \Delta p(n)=p(n+1)-p(n)}$ 

求出各阶差分：${\displaystyle p(n)=a+(n-1)d,\Delta p(n)=d}$ 

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

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

如果 -1 < r < 1，那么其无穷级数为^[1]

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

如果r在上述范围之外，则该级数不是发散级数就是交错级数。

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