裂項和

（重定向自裂项消元法

${\displaystyle \sum _{i=1}^{n}(a_{i}-a_{i+1})=(a_{1}-a_{2})+(a_{2}-a_{3})+\ldots +(a_{n}-a_{n+1})=a_{1}-a_{n+1}.}$

${\displaystyle \prod _{i=1}^{n}{\frac {a_{i}}{a_{i+1}}}={\frac {a_{1}}{a_{n+1}}}}$

可以用來裂项求和的數學式

${\displaystyle {\frac {1}{a(a+b)}}={\frac {1}{b}}\left({\frac {1}{a}}-{\frac {1}{a+b}}\right)}$

${\displaystyle a^{k}={\frac {1}{a-1}}(a^{k+1}-a^{k})}$

${\displaystyle \cos kx={\frac {1}{2\sin {\frac {x}{2}}}}\left[\sin \left(k+{\frac {1}{2}}\right)x-\sin \left(k-{\frac {1}{2}}\right)x\right]}$

${\displaystyle \sin kx={\frac {1}{2\sin {\frac {x}{2}}}}\left[\cos \left(k-{\frac {1}{2}}\right)x-\cos \left(k+{\frac {1}{2}}\right)x\right]}$ 三角恒等式[1]

${\displaystyle C_{k}^{n}=C_{k}^{n-1}+C_{k-1}^{n-1}}$ 帕斯卡法則

${\displaystyle {\frac {1}{C_{k}^{n}}}={\frac {n+1}{n+2}}\left({\frac {1}{C_{k}^{n+1}}}+{\frac {1}{C_{k+1}^{n+1}}}\right)}$ [2]

求和类型

一般求和

${\displaystyle {\frac {1}{1\cdot 2}}+{\frac {1}{2\cdot 3}}+{\frac {1}{3\cdot 4}}+\cdots +{\frac {1}{n\cdot (n+1)}}=\sum _{k=1}^{n}{\frac {1}{k(k+1)}}=\sum _{k=1}^{n}{\frac {1}{k}}-{\frac {1}{k+1}}=({\frac {1}{1}}-{\frac {1}{2}})+({\frac {1}{2}}-{\frac {1}{3}})+({\frac {1}{3}}-{\frac {1}{4}})+\cdots +({\frac {1}{n}}-{\frac {1}{n+1}})=1-{\frac {1}{n+1}}}$

交错求和

${\displaystyle \sum _{k=1}^{2n}{\frac {(-1)^{k-1}}{C_{2n}^{k}}}={\frac {2n+1}{2n+2}}\sum _{k=1}^{2n}(-1)^{k-1}\left({\frac {1}{C_{2n+1}^{k}}}+{\frac {1}{C_{2n+1}^{k+1}}}\right)={\frac {2n+1}{2n+2}}\left[{\frac {1}{C_{2n+1}^{1}}}+{\frac {(-1)^{2n-1}}{C_{2n+1}^{2n+1}}}\right]={\frac {2n+1}{2n+2}}\left({\frac {-2n}{2n+1}}\right)={\frac {-n}{n+1}}}$

誤用

${\displaystyle 0=\sum _{n=1}^{\infty }0=\sum _{n=1}^{\infty }(1-1)=1+\sum _{n=1}^{\infty }(-1+1)=1\,}$

${\displaystyle \sum _{n=1}^{N}{\frac {1}{n(n+1)}}=\sum _{n=1}^{N}{\frac {1}{n}}-{\frac {1}{n+1}}\,}$
${\displaystyle =\left(1-{\frac {1}{2}}\right)+\left({\frac {1}{2}}-{\frac {1}{3}}\right)+\cdots +\left({\frac {1}{N}}-{\frac {1}{N+1}}\right)\,}$
${\displaystyle =1+\left(-{\frac {1}{2}}+{\frac {1}{2}}\right)+\left(-{\frac {1}{3}}+{\frac {1}{3}}\right)+\cdots +\left(-{\frac {1}{N}}+{\frac {1}{N}}\right)-{\frac {1}{N+1}}\,}$
${\displaystyle =1-{\frac {1}{N+1}}\to 1\ \mathrm {as} \ N\to \infty .\,}$

例子：三角函數

${\displaystyle \sum _{n=1}^{N}\sin \left(n\right)=\sum _{n=1}^{N}{\frac {1}{2}}\csc \left({\frac {1}{2}}\right)\left[2\sin \left({\frac {1}{2}}\right)\sin \left(n\right)\right]}$
${\displaystyle ={\frac {1}{2}}\csc \left({\frac {1}{2}}\right)\sum _{n=1}^{N}\left[\cos \left({\frac {2n-1}{2}}\right)-\cos \left({\frac {2n+1}{2}}\right)\right]}$
${\displaystyle ={\frac {1}{2}}\csc \left({\frac {1}{2}}\right)\left[\cos \left({\frac {1}{2}}\right)-\cos \left({\frac {2N+1}{2}}\right)\right]}$

參考資料

1. ^ 唐秀农. 裂项法求和的一般原理和法则. 数学教学通讯. 2013, (9) [2014-06-17]. （原始内容存档于2014-07-14）.
2. ^ 及万会 张来萍 杨春艳. 封闭形和式初步.