# 二項式定理

${\displaystyle (x+y)^{4}\;=\;x^{4}\,+\,4x^{3}y\,+\,6x^{2}y^{2}\,+\,4xy^{3}\,+\,y^{4}.}$

axbyc 中的係數a被稱為二項式係數，記作 ${\displaystyle {\tbinom {n}{b}}}$${\displaystyle {\tbinom {n}{c}}}$（二者值相等）。二項式定理可以推廣到任意實數次冪，即廣義二項式定理[2]

## 定理的陳述

${\displaystyle (x+y)^{n}={n \choose 0}x^{n}y^{0}+{n \choose 1}x^{n-1}y^{1}+{n \choose 2}x^{n-2}y^{2}+\cdots +{n \choose n-1}x^{1}y^{n-1}+{n \choose n}x^{0}y^{n},}$

${\displaystyle (x+y)^{n}=\sum _{k=0}^{n}{n \choose k}x^{n-k}y^{k}=\sum _{k=0}^{n}{n \choose k}x^{k}y^{n-k}.}$

${\displaystyle (1+x)^{n}={n \choose 0}x^{0}+{n \choose 1}x^{1}+{n \choose 2}x^{2}+\cdots +{n \choose {n-1}}x^{n-1}+{n \choose n}x^{n},}$

${\displaystyle (1+x)^{n}=\sum _{k=0}^{n}{n \choose k}x^{k}.}$

## 證明

### 數學歸納法

${\displaystyle n=1}$

${\displaystyle (a+b)^{1}=\sum _{k=0}^{1}{1 \choose k}a^{1-k}b^{k}={1 \choose 0}a^{1}b^{0}+{1 \choose 1}a^{0}b^{1}=a+b}$

${\displaystyle (a+b)^{m+1}}$ ${\displaystyle =}$ ${\displaystyle a(a+b)^{m}+b(a+b)^{m}}$
${\displaystyle =}$ ${\displaystyle a\sum _{k=0}^{m}{m \choose k}a^{m-k}b^{k}+b\sum _{j=0}^{m}{m \choose j}a^{m-j}b^{j}}$
${\displaystyle =}$ ${\displaystyle \sum _{k=0}^{m}{m \choose k}a^{m-k+1}b^{k}+\sum _{j=0}^{m}{m \choose j}a^{m-j}b^{j+1}}$${\displaystyle a}$${\displaystyle b}$
${\displaystyle =}$ ${\displaystyle a^{m+1}+\sum _{k=1}^{m}{m \choose k}a^{m-k+1}b^{k}+\sum _{j=0}^{m}{m \choose j}a^{m-j}b^{j+1}}$ 取出${\displaystyle k=0}$的項
${\displaystyle =}$ ${\displaystyle a^{m+1}+\sum _{k=1}^{m}{m \choose k}a^{m-k+1}b^{k}+\sum _{k=1}^{m+1}{m \choose k-1}a^{m-k+1}b^{k}}$${\displaystyle j=k-1}$
${\displaystyle =}$ ${\displaystyle a^{m+1}+\sum _{k=1}^{m}{m \choose k}a^{m-k+1}b^{k}+\sum _{k=1}^{m}{m \choose k-1}a^{m+1-k}b^{k}+b^{m+1}}$ 取出${\displaystyle k=m+1}$
${\displaystyle =}$ ${\displaystyle a^{m+1}+b^{m+1}+\sum _{k=1}^{m}\left[{m \choose k}+{m \choose k-1}\right]a^{m+1-k}b^{k}}$ 兩者加起
${\displaystyle =}$ ${\displaystyle a^{m+1}+b^{m+1}+\sum _{k=1}^{m}{m+1 \choose k}a^{m+1-k}b^{k}}$ 套用帕斯卡法則
${\displaystyle =}$ ${\displaystyle \sum _{k=0}^{m+1}{m+1 \choose k}a^{m+1-k}b^{k}}$

### 一般形式的證明

${\displaystyle f(x)=(1+x)^{a}}$, ${\displaystyle g(x)=\sum _{k=0}^{\infty }{a \choose k}x^{k}}$. 注意只有當 ${\displaystyle |x|<1}$時上述兩個函數才收斂

• 首先證明 ${\displaystyle f(x)}$收斂於1. 這裡省略
• 之後, 易得${\displaystyle f(x)}$滿足微分方程式: ${\displaystyle (1+x)f'(x)=af(x)}$. 用求導的一般方法就能得到這個結論, 這裡省略
• 再證明 ${\displaystyle g(x)}$亦滿足上述微分方程式:

${\displaystyle g(x)=1+\sum _{k=1}^{\infty }{a \choose k}x^{k}}$

{\displaystyle {\begin{aligned}g'(x)&=\sum _{k=1}^{\infty }{a \choose k}kx^{k-1}\\&=\sum _{k=0}^{\infty }{a \choose {k+1}}(k+1)x^{k}\\&=\sum _{k=0}^{\infty }{a \choose k}(a-k)x^{k}\\\end{aligned}}}

{\displaystyle {\begin{aligned}{a \choose {k+1}}(k+1)&={\frac {(a)(a-1)\cdots (a-k+1)(a-k)}{(k+1)!}}(k+1)\\&={\frac {(a)(a-1)\cdots (a-k+1)(a-k)}{k!}}\\&={a \choose k}(a-k)\end{aligned}}}

${\displaystyle =\sum _{k=0}^{\infty }{a \choose k}(a-k)x^{k}+\sum _{k=1}^{\infty }{a \choose k}kx^{k}}$
${\displaystyle =\sum _{k=0}^{\infty }{a \choose k}(a-k)x^{k}+\sum _{k=0}^{\infty }{a \choose k}kx^{k}}$
${\displaystyle =\sum _{k=0}^{\infty }{a \choose k}x^{k}(a-k+k)}$
${\displaystyle =a\sum _{k=0}^{\infty }{a \choose k}x^{k}}$
${\displaystyle =a\cdot g(x)}$

${\displaystyle {\frac {g'(x)}{g(x)}}={\frac {a}{1+x}}}$
${\displaystyle \because {\frac {f'(x)}{f(x)}}={\frac {a}{1+x}}}$
${\displaystyle \therefore {\frac {f'(x)}{f(x)}}={\frac {g'(x)}{g(x)}}}$
${\displaystyle g'(x)f(x)=f'(x)g(x)}$
• 根據除法定則${\displaystyle {\frac {d}{dx}}\left({\frac {g(x)}{f(x)}}\right)={\frac {g'(x)f(x)-f'(x)g(x)}{(f(x))^{2}}}=0}$
${\displaystyle {\frac {g(x)}{f(x)}}={\frac {g(0)}{f(0)}}=1}$
${\displaystyle f(x)=g(x)}$

## 應用

### 多倍角恆等式

${\displaystyle \cos \left(nx\right)+i\sin \left(nx\right)=\left(\cos x+i\sin x\right)^{n}.\,}$

${\displaystyle \left(\cos x+i\sin x\right)^{2}=\cos ^{2}x+2i\cos x\sin x-\sin ^{2}x,}$

${\displaystyle \cos(2x)=\cos ^{2}x-\sin ^{2}x\quad {\text{and}}\quad \sin(2x)=2\cos x\sin x,}$

${\displaystyle \left(\cos x+i\sin x\right)^{3}=\cos ^{3}x+3i\cos ^{2}x\sin x-3\cos x\sin ^{2}x-i\sin ^{3}x,}$

${\displaystyle \cos(3x)=\cos ^{3}x-3\cos x\sin ^{2}x\quad {\text{and}}\quad \sin(3x)=3\cos ^{2}x\sin x-\sin ^{3}x.}$

${\displaystyle \cos(nx)=\sum _{k{\text{ even}}}(-1)^{\frac {k}{2}}{n \choose k}\cos ^{n-k}x\sin ^{k}x}$

${\displaystyle \sin(nx)=\sum _{k{\text{ odd}}}(-1)^{\frac {k-1}{2}}{n \choose k}\cos ^{n-k}x\sin ^{k}x.}$

### e級數

${\displaystyle e=\lim _{n\to \infty }\left(1+{\frac {1}{n}}\right)^{n}.}$

${\displaystyle \left(1+{\frac {1}{n}}\right)^{n}=1+{n \choose 1}{\frac {1}{n}}+{n \choose 2}{\frac {1}{n^{2}}}+{n \choose 3}{\frac {1}{n^{3}}}+\cdots +{n \choose n}{\frac {1}{n^{n}}}.}$

k項之總和為

${\displaystyle {n \choose k}{\frac {1}{n^{k}}}\;=\;{\frac {1}{k!}}\cdot {\frac {n(n-1)(n-2)\cdots (n-k+1)}{n^{k}}}}$

${\displaystyle \lim _{n\to \infty }{n \choose k}{\frac {1}{n^{k}}}={\frac {1}{k!}}.}$

${\displaystyle e=\sum _{k=0}^{\infty }{\frac {1}{k!}}={\frac {1}{0!}}+{\frac {1}{1!}}+{\frac {1}{2!}}+{\frac {1}{3!}}+\cdots .}$

## 推廣

${\displaystyle (x+y)^{\alpha }=\sum _{k=0}^{\infty }{\alpha \choose k}x^{\alpha -k}y^{k}}$。其中${\displaystyle {\alpha \choose k}={\frac {\alpha (\alpha -1)...(\alpha -k+1)}{k!}}={\frac {(\alpha )_{k}}{k!}}}$

### 多項式展開

${\displaystyle \left(x_{1}+x_{2}+...+x_{n}\right)^{k}=\sum _{\alpha _{1}+\alpha _{2}+...+\alpha _{n}=k}{\frac {k!}{\alpha _{1}!...\alpha _{n}!}}x_{1}^{\alpha _{1}}...x_{n}^{\alpha _{n}}}$.

${\displaystyle \left(x_{1}+x_{2}+...+x_{n}\right)^{k}}$
${\displaystyle =}$ ${\displaystyle ((x_{1}+x_{2}+...+x_{n-1})+x_{n})^{k}}$
${\displaystyle =}$ ${\displaystyle \sum _{\alpha _{n}=0}^{k}{\frac {k!}{\alpha _{n}!\left(k-\alpha _{n}\right)!}}\left(x_{1}+x_{2}+...+x_{n-1}\right)^{k-\alpha _{n}}x_{n}^{\alpha _{n}}}$
${\displaystyle =}$ ${\displaystyle \sum _{\alpha _{n}=0}^{k}{\frac {k!}{\alpha _{n}!\left(k-\alpha _{n}\right)!}}\sum _{\alpha _{1}+\alpha _{2}+...+\alpha _{n-1}=k-\alpha _{n}}{\frac {\left(k-\alpha _{n}\right)!}{\alpha _{1}!...\alpha _{n-1}!}}x_{1}^{\alpha _{1}}...x_{n-1}^{\alpha _{n-1}}x_{n}^{\alpha _{n}}}$
${\displaystyle =}$ ${\displaystyle \sum _{\alpha _{1}+\alpha _{2}+...+\alpha _{n}=k}{\frac {k!}{\alpha _{1}!...\alpha _{n}!}}x_{1}^{\alpha _{1}}...x_{n}^{\alpha _{n}}}$證畢

## 參考文獻

1. ^ Binomial Expansions - leeds uk
2. ^ Roman, Steven "The Umbral Calculus", Dover Publications, 2005, ISBN 0-486-44129-3
3. ^ Devlin, Keith, The Unfinished Game: Pascal, Fermat, and the Seventeenth-Century Letter that Made the World Modern, Basic Books; 1 edition (2008), ISBN 978-0-465-00910-7, p. 24.
4. ^ Binomial Theorem - wolfram mathworld
5. ^ The Story of the Binomial Theorem by J. L. Coolidge, The American Mathematical Monthly 56:3 (1949), pp. 147–157
6. O'Connor, John J.; Robertson, Edmund F., Abu Bekr ibn Muhammad ibn al-Husayn Al-Karaji, MacTutor History of Mathematics archive （英語）
7. ^ Landau, James A. Historia Matematica Mailing List Archive: Re: [HM] Pascal's Triangle (mailing list email). Archives of Historia Matematica. 1999-05-08 [2007-04-13]. 參數|title=值左起第1位存在刪除符 (幫助)
8. ^ Bourbaki: History of mathematics
9. ^ The Geometry of the Binomial Theorem The Geometry of the Binomial Theorem - Math Awareness
10. ^ Błaszczyk, Piotr; Katz, Mikhail; Sherry, David, Ten misconceptions from the history of analysis and their debunking, Foundations of Science, 2012, arXiv:1202.4153, doi:10.1007/s10699-012-9285-8
11. ^ Multiple-Angle Formulas - MathWorld
12. ^ The Constant e - NDE/NDT Resource Center
13. ^ Series - NTEC
14. ^
15. ^ 多項式定理的新證明及其展開 - 佛山科學技術學院信息科學與數學系
16. ^ Hazewinkel, Michiel (編), Multinomial coefficient, 數學百科全書, Springer, 2001, ISBN 978-1-55608-010-4

## 參考書目

• Bag, Amulya Kumar. Binomial theorem in ancient India. Indian J. History Sci. 1966, 1 (1): 68–74.
• Barth, Nils R. (November 2004). "Computing Cavalieri's Quadrature Formula by a Symmetry of the n-Cube". The American Mathematical Monthly (Mathematical Association of America) 111 (9): 811–813. doi:10.2307/4145193. ISSN 0002-9890. JSTOR 4145193, author's copy, further remarks and resources
• Graham, Ronald; Knuth, Donald; Patashnik, Oren. (5) Binomial Coefficients. Concrete Mathematics 2nd. Addison Wesley. 1994: 153–256. ISBN 0-201-55802-5. OCLC 17649857.