# 代數/本書課文/求和/裂項法

## 一般裂項

${\displaystyle x_{i}=y_{i+1}-y_{i}}$

${\displaystyle \sum _{i=a}^{b}x_{i}=\sum _{i=a}^{b}(y_{i+1}-y_{i})=y_{b+1}-y_{b}+y_{b}-y_{b-1}+\cdots +y_{a+1}-y_{a}=y_{b+1}-y_{a}}$

${\displaystyle x_{i}=y_{i}-y_{i+1}}$

${\displaystyle \sum _{i=a}^{b}x_{i}=\sum _{i=a}^{b}(y_{i}-y_{i+1})=y_{a}-y_{a+1}+y_{a+1}-y_{a+2}+\cdots +y_{b}-y_{b+1}=y_{a}-y_{b+1}}$}}

${\displaystyle \Delta y_{i}=y_{i+1}-y_{i}}$，以上求和可以寫成：${\displaystyle \sum _{i=a}^{b}\Delta y_{i}=y_{b+1}-y_{a},\sum _{i=a}^{b}-\Delta y_{i}=y_{a}-y_{b+1}}$

${\displaystyle \sum _{k=1}^{n}{\frac {1}{k(k+1)}}=\sum _{k=1}^{n}({\frac {1}{k}}-{\frac {1}{k+1}})=1-{\frac {1}{n+1}}}$
${\displaystyle \sum _{k=1}^{n}q^{k-1}=\sum _{k=1}^{n}{\frac {q^{k}-q^{k-1}}{q-1}}={\frac {q^{n}-1}{q-1}}}$（等比數列求和）
${\displaystyle \sum _{k=a}^{b}C_{k}^{m}=\sum _{k=a}^{b}(C_{k+1}^{m+1}-C_{k}^{m+1})=C_{b+1}^{m+1}-C_{a}^{m+1}}$

## 隔幾項裂項

${\displaystyle x_{i}=y_{i+k}-y_{i}=\sum _{j=0}^{k-1}(y_{i+j+1}-y_{i+j})=\sum _{j=0}^{k-1}\Delta y_{i+j}}$

${\displaystyle \sum _{i=a}^{b}x_{i}=\sum _{j=0}^{k-1}\sum _{i=a}^{b}\Delta y_{i+j}=\sum _{j=0}^{k-1}(y_{b+1+j}-y_{a+j})}$

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

## 和裂項

${\displaystyle x_{i}=y_{i+1}+y_{i}}$

${\displaystyle \sum _{i=a}^{b}(-1)^{i}x_{i}=\sum _{i=a}^{b}(-1)^{i}(y_{i+1}+y_{i})=-\sum _{i=a}^{b}[(-1)^{i+1}y_{i+1}-(-1)^{i}y_{i}]}$
${\displaystyle =-\sum _{i=a}^{b}\Delta [(-1)^{i}y_{i}]=(-1)^{a}y_{a}-(-1)^{b+1}y_{b+1}}$

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

## 待定裂項法

 例子：差比數列求和 待定係數s,t使得差比數列可以裂項： ${\displaystyle [a+(k-1)d]r^{k-1}=(sk+t)r^{k}-[s(k-1)+t]r^{k-1}}$ ${\displaystyle \sum _{k=1}^{n}[a+(k-1)d]r^{k-1}=(sn+t)r^{n}-t}$ 求出待定係數s,t關於a,d,r的表達式： ${\displaystyle dk+a-d=s(r-1)k+(r-1)t+s}$ ${\displaystyle s={\frac {d}{r-1}},t={\frac {a-d}{r-1}}-{\frac {d}{(r-1)^{2}}}}$ ${\displaystyle \sum _{k=1}^{n}[a+(k-1)d]r^{k-1}=[{\frac {d}{r-1}}n+{\frac {a-d}{r-1}}-{\frac {d}{(r-1)^{2}}}]r^{n}-[{\frac {a-d}{r-1}}-{\frac {d}{(r-1)^{2}}}]}$[1]

${\displaystyle \sum _{k=1}^{n}p(k)q^{k-1}=f(n)q^{n}-f(0)}$

 例子：求${\displaystyle \displaystyle \sum _{k=1}^{n}(A+Bk+Ck^{2})q^{k-1}}$ ${\displaystyle (A+Bk+Ck^{2})q^{k-1}=(D+Ek+Fk^{2})q^{k}-[D+E(k-1)+F(k-1)^{2}]q^{k-1}}$ ${\displaystyle A+Bk+Ck^{2}=(qD+qEk+qFk^{2})-[D+E(k-1)+F(k^{2}-2k+1)]}$ ${\displaystyle A+Bk+Ck^{2}=[(q-1)D+E-F]+[(q-1)E+2F]k+(q-1)Fk^{2}}$ ${\displaystyle \displaystyle F={\frac {C}{q-1}},E={\frac {B}{q-1}}-{\frac {2C}{(q-1)^{2}}},D={\frac {A}{q-1}}-{\frac {B-C}{(q-1)^{2}}}+{\frac {2C}{(q-1)^{3}}}}$ ${\displaystyle f(n)=[{\frac {A}{q-1}}-{\frac {B-C}{(q-1)^{2}}}+{\frac {2C}{(q-1)^{3}}}]+[{\frac {B}{q-1}}-{\frac {2C}{(q-1)^{2}}}]n+{\frac {C}{q-1}}n^{2}}$ ${\displaystyle \sum _{k=1}^{n}(A+Bk+Ck^{2})q^{k-1}=\{[{\frac {A}{q-1}}-{\frac {B-C}{(q-1)^{2}}}+{\frac {2C}{(q-1)^{3}}}]+[{\frac {B}{q-1}}-{\frac {2C}{(q-1)^{2}}}]n+{\frac {C}{q-1}}n^{2}\}q^{n}-[{\frac {A}{q-1}}-{\frac {B-C}{(q-1)^{2}}}+{\frac {2C}{(q-1)^{3}}}]}$

 例子：等差數列求和 待定係數A,B,C使得等差數列可以裂項： ${\displaystyle a+(k-1)d=A+Bk+Ck^{2}-[A+B(k-1)+C(k-1)^{2}]}$ ${\displaystyle \sum _{k=1}^{n}[a+(k-1)d]=A+Bn+Cn^{2}-A=Bn+Cn^{2}}$ 求出待定係數B,C關於a,d的表達式： ${\displaystyle a-d+dk=B-C+2Ck}$ ${\displaystyle C={\frac {d}{2}},B=a-{\frac {d}{2}}}$ ${\displaystyle \sum _{k=1}^{n}[a+(k-1)d]=(a-{\frac {d}{2}})n+{\frac {d}{2}}n^{2}=an+d{\frac {n(n-1)}{2}}}$

## 參考資料

1. 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).