等差-等比数列

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。

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

对等比数列和两边求导：[3]

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

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

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：

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

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

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

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

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

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

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

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