查看“︁代數/本書課文/求和/裂項法”︁的源代码

若數列存在一個裂項變換，則可對此數列使用裂項法求和。

==一般裂項==
若<math>x_i=y_{i+1}-y_i</math>
:<math>\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</math><br/>
若<math>x_i=y_i-y_{i+1}</math>
:<math>\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}</math>}}

記<math>\Delta y_i=y_{i+1}-y_i</math>，以上求和可以寫成：<math>\sum_{i=a}^b \Delta y_i=y_{b+1}-y_a,\sum_{i=a}^b -\Delta y_i=y_a-y_{b+1}</math>

{{Example|
<math>\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}</math><br/>
<math>\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}</math>（等比數列求和）<br/>
<math>\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}</math>
}}

==隔幾項裂項==
若<math>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}</math>
:<math>\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})</math>

{{Example|
<math>\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})</math>
}}

==和裂項==
若<math>x_i=y_{i+1}+y_i</math>
:<math>\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]</math>
:<math>=-\sum_{i=a}^b \Delta [(-1)^i y_i]=(-1)^a y_a-(-1)^{b+1}y_{b+1}</math>

{{Example|
<math>\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}</math>
}}

==待定裂項法==
{{ExampleRobox|title=例子：差比數列求和}}
待定係數s,t使得差比數列可以裂項：<br/>
<math>[a+(k-1)d]r^{k-1}=(sk+t)r^k-[s(k-1)+t]r^{k-1}</math><br/>
<math>\sum_{k=1}^n [a+(k-1)d]r^{k-1}=(sn+t)r^n-t</math><br/>
求出待定係數s,t關於a,d,r的表達式：<br/>
<math>dk+a-d=s(r-1)k+(r-1)t+s</math><br/>
<math>s=\frac{d}{r-1},t=\frac{a-d}{r-1}-\frac{d}{(r-1)^2}</math><br/>
<math>\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}]</math><ref>{{cite journal|author=郑良|year=2012|title=差比型数列前n项和的求解方法——裂项法|journal=中学生数学|issue=3|url=http://www.cnki.com.cn/Article/CJFDTOTAL-ZXSS201203012.htm}}</ref>
{{Robox/Close}}

對於多項式公比求和<math>\sum_{k=1}^n p(k)q^{k-1}</math>，對數列做裂項：<math>p(k)q^{k-1}=f(k)q^k-f(k-1)q^{k-1}</math>

其中若<math>p(k)</math>是m阶階多項式，則<math>f(k)</math>是m階多項式，<math>f(k)</math>用待定係數法求出來。

<math>\sum_{k=1}^n p(k)q^{k-1}=f(n)q^n-f(0)</math>

{{ExampleRobox|title=例子：求<math>\displaystyle\sum_{k=1}^n (A+Bk+Ck^2)q^{k-1}</math>}}

<math>(A+Bk+Ck^2)q^{k-1}=(D+Ek+Fk^2)q^k-[D+E(k-1)+F(k-1)^2]q^{k-1}</math>

<math>A+Bk+Ck^2=(qD+qEk+qFk^2)-[D+E(k-1)+F(k^2-2k+1)]</math>

<math>A+Bk+Ck^2=[(q-1)D+E-F]+[(q-1)E+2F]k+(q-1)Fk^2</math>

<math>\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}

</math>

<math>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</math>

<math>\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}]</math>

{{Robox/Close}}

對於多項式求和<math>\sum_{k=1}^n p(k)</math>，對數列做裂項：<math>p(k)=f(k)-f(k-1)</math>

其中若<math>p(k)</math>是m阶階多項式，則<math>f(k)</math>是m+1階多項式，<math>f(k)</math>用待定係數法求出來。

{{ExampleRobox|title=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：<br/>
<math>a+(k-1)d=A+Bk+Ck^2-[A+B(k-1)+C(k-1)^2]</math><br/>
<math>\sum_{k=1}^n [a+(k-1)d]=A+Bn+Cn^2-A=Bn+Cn^2</math><br/>
求出待定係數B,C關於a,d的表達式：
<math>a-d+dk=B-C+2Ck</math><br/>
<math>C=\frac{d}{2},B=a-\frac{d}{2}</math><br/>
<math>\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}</math><br/>
{{Robox/Close}}

==參考資料==
{{reflist}}