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裂项求和（Telescoping sum）是一個非正式的用語，指一種用來計算級數的技巧：每項可以分拆，令上一項和下一項的某部分互相抵消，剩下頭尾的項需要計算，從而求得級數和。

∑i=1n(ai−ai+1)=(a1−a2)+(a2−a3)+…+(an−an+1)=a1−an+1.{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}.}

裂項積（Telescoping product）也是差不多的概念：

∏i=1naiai+1=a1an+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

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

ak=1a−1(ak+1−ak){displaystyle a^{k}={frac {1}{a-1}}(a^{k+1}-a^{k})}

cos⁡kx=12sin⁡x2[sin⁡(k+12)x−sin⁡(k−12)x]{displaystyle cos kx={frac {1}{2sin {frac {x}{2}}}}left[sin left(k+{frac {1}{2}}right)x-sin left(k-{frac {1}{2}}right)xright]}

sin⁡kx=12sin⁡x2[cos⁡(k−12)x−cos⁡(k−12)x]{displaystyle sin kx={frac {1}{2sin {frac {x}{2}}}}left[cos left(k-{frac {1}{2}}right)x-cos left(k-{frac {1}{2}}right)xright]}（三角恒等式）^[1]

Ckn=Ckn−1+Ck−1n−1{displaystyle C_{k}^{n}=C_{k}^{n-1}+C_{k-1}^{n-1}}（帕斯卡法則）

1Ckn=n+1n+2(1Ckn+1+1Ck+1n+1){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

若有ak=bk−bk+1{displaystyle a_{k}=b_{k}-b_{k+1}}，则∑k=mnak=bm−bn+1{displaystyle sum _{k=m}^{n}a_{k}=b_{m}-b_{n+1}}

11⋅2+12⋅3+13⋅4+⋯+1n⋅(n+1)=∑k=1n1k(k+1)=∑k=1n1k−1k+1=(11−12)+(12−13)+(13−14)+⋯+(1n−1n+1)=1−1n+1{displaystyle {frac {1}{1cdot 2}}+{frac {1}{2cdot 3}}+{frac {1}{3cdot 4}}+cdots +{frac {1}{ncdot (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

若有ak=bk+bk+1{displaystyle a_{k}=b_{k}+b_{k+1}}，则∑k=mn(−1)kak=(−1)mbm+(−1)n+1bn+1{displaystyle sum _{k=m}^{n}(-1)^{k}a_{k}=(-1)^{m}b_{m}+(-1)^{n+1}b_{n+1}}

∑k=12n(−1)k−1C2nk=2n+12n+2∑k=12n(−1)k−1(1C2n+1k+1C2n+1k+1)=2n+12n+2[1C2n+11+(−1)2n−1C2n+12n+1]=2n+12n+2(−2n2n+1)=−nn+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}}}

誤用

0=∑n=1∞0=∑n=1∞(1−1)=1+∑n=1∞(−1+1)=1{displaystyle 0=sum _{n=1}^{infty }0=sum _{n=1}^{infty }(1-1)=1+sum _{n=1}^{infty }(-1+1)=1,}

這是錯誤的。將每項重組的方法只適用於獨立的項趨近0。

防止這種錯誤，可以先求首N項的值，然後取N趨近無限的值。

∑n=1N1n(n+1)=∑n=1N1n−1(n+1){displaystyle sum _{n=1}^{N}{frac {1}{n(n+1)}}=sum _{n=1}^{N}{frac {1}{n}}-{frac {1}{(n+1)}},}

=(1−12)+(12−13)+⋯+(1N−1N+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),}

=1+(−12+12)+(−13+13)+⋯+(−1N+1N)−1N+1{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}},}

=1−1N+1→1 as N→∞.{displaystyle =1-{frac {1}{N+1}}to 1 mathrm {as} Nto infty .,}

例子：三角函數

∑n=1Nsin⁡(n)=∑n=1N12csc⁡(12)[2sin⁡(12)sin⁡(n)]{displaystyle sum _{n=1}^{N}sin left(nright)=sum _{n=1}^{N}{frac {1}{2}}csc left({frac {1}{2}}right)left[2sin left({frac {1}{2}}right)sin left(nright)right]}

=12csc⁡(12)∑n=1N[cos⁡(2n−12)−cos⁡(2n+12)]{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]}

=12csc⁡(12)[cos⁡(12)−cos⁡(2N+12)]{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

外部連結

• https://web.archive.org/web/20060902104046/http://www.math.wisc.edu/~rhoades/Notes/buFall2005putnam.pdf