二项式系数 出现在杨辉三角 （帕斯卡三角）中。除边缘的数字外，其他每一个数都为其上方两数之和。二项式定理 （英語：Binomial theorem ）描述了二项式 的幂 的代数 展开。根据该定理，可以将两个数之和的整数次幂诸如
(
x
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n
{\displaystyle (x+y)^{n}}
a
x
b
y
c
{\displaystyle ax^{b}y^{c}}
恒等式 ，其中
b
{\displaystyle b}
c
{\displaystyle c}
b
+
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=
n
{\displaystyle b+c=n}
a
{\displaystyle a}
n
{\displaystyle n}
b
{\displaystyle b}
[ 1]
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4
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x
2
y
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y
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+
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4
.
{\displaystyle (x+y)^{4}\;=\;x^{4}\,+\,4x^{3}y\,+\,6x^{2}y^{2}\,+\,4xy^{3}\,+\,y^{4}.}
a
x
b
y
c
{\displaystyle ax^{b}y^{c}}
a
{\displaystyle a}
二项式系数 ，记作
(
n
b
)
{\displaystyle {\tbinom {n}{b}}}
(
n
c
)
{\displaystyle {\tbinom {n}{c}}}
广义二项式定理 [ 2]
二项式系数的三角形排列通常被认为是法国数学家布莱兹·帕斯卡 的贡献，他在17世纪描述了这一现象[ 3] 欧几里得 于公元前4世纪提到了指数为2的情况[ 4] [ 5] 青目 探讨了更高阶的情况。帕斯卡三角形 的雏形于10世纪由印度数学家大力羅摩 发现。在同一时期，波斯数学家卡拉吉 [ 6] 歐瑪爾·海亞姆 得到了更为普遍的二项式定理的形式。13世纪，中国数学家杨辉 也得到了类似的结果[ 7] 卡拉吉 数学归纳法 的原始形式给出了二项式定理和帕斯卡三角形 (巴斯卡三角形 )的有关证明[ 6] 艾萨克·牛顿 勋爵将二项式定理的系数推广到有理数 [ 8]
根据此定理，可以将
x
+
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{\displaystyle x+y}
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{\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},}
其中每个
(
n
k
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{\displaystyle {\tbinom {n}{k}}}
二项式系数 的特定正整数，其等於
n
!
k
!
(
n
−
k
)
!
{\displaystyle {\frac {n!}{k!(n-k)!}}}
二项式公式 或二项恒等式 。使用求和符号 ，可以把它写作
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n
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{\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}.}
后面的表达式只是将根据
x
{\displaystyle x}
y
{\displaystyle y}
y
{\displaystyle y}
变量 。在这种形式中，公式写作
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{\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},}
或者等价地
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n
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0
n
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x
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{\displaystyle (1+x)^{n}=\sum _{k=0}^{n}{n \choose k}x^{k}.}
对直到四次幂的二项式的可视化 对于正值
a
{\displaystyle a}
b
{\displaystyle b}
n
=
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{\displaystyle n=2}
a
+
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{\displaystyle a+b}
a
{\displaystyle a}
b
{\displaystyle b}
a
{\displaystyle a}
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{\displaystyle b}
n
=
3
{\displaystyle n=3}
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+
b
{\displaystyle a+b}
a
{\displaystyle a}
b
{\displaystyle b}
a
×
a
×
b
{\displaystyle a\times a\times b}
a
×
b
×
b
{\displaystyle a\times b\times b}
在微积分 中，此图解也给出导数
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′
=
n
x
n
−
1
{\displaystyle (x^{n})'=nx^{n-1}}
[ 9]
a
=
x
{\displaystyle a=x}
b
=
Δ
x
{\displaystyle b=\Delta x}
b
{\displaystyle b}
a
{\displaystyle a}
无穷小量 改变，则此图解将无穷小量改变，显示为
n
{\displaystyle n}
超立方体
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x
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n
{\displaystyle (x+\Delta x)^{n}}
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{\displaystyle (x+\Delta x)^{n}=x^{n}+nx^{n-1}\Delta x+{\tbinom {n}{2}}x^{n-2}(\Delta x)^{2}+\cdots .}
其中（针对
Δ
x
{\displaystyle \Delta x}
n
x
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−
1
{\displaystyle nx^{n-1}}
差商 的导数 定义 并取极限，意味着忽略高阶项
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2
{\displaystyle (\Delta x)^{2}}
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′
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x
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{\displaystyle (x^{n})'=nx^{n-1}}
微积分基本定理 ，则得到卡瓦列里求积公式 ：
∫
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n
−
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d
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x
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{\displaystyle \textstyle {\int x^{n-1}\,dx={\tfrac {1}{n}}x^{n}}}
當
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{\displaystyle n=1}
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{\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}
假設二项展开式在
n
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{\displaystyle n=m}
n
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{\displaystyle n=m+1}
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( 將
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取 出
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的 項
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設
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取 出
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[
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+
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]
a
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兩 者 加 起
=
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套 用 帕 斯 卡 法 則
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{\displaystyle {\begin{aligned}(a+b)^{m+1}&=a(a+b)^{m}+b(a+b)^{m}\\&=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}\\&=\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}\ \ \ \ \ {\text{( 將 }}a{\text{, }}b{\text{) }}\\&=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}\ \ \ \ \ {\text{ 取 出 }}k=0{\text{ 的 項 }}\\&=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}\ \ \ \ \ {\text{設 }}j=k-1\\&=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}\ \ \ \ \ {\text{取 出 }}k=m+1{\text{項 }}\\&=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}\ \ \ \ \ {\text{兩 者 加 起 }}\\&=a^{m+1}+b^{m+1}+\sum _{k=1}^{m}{m+1 \choose k}a^{m+1-k}b^{k}\ \ \ \ \ {\text{套 用 帕 斯 卡 法 則 }}\\&=\sum _{k=0}^{m+1}{m+1 \choose k}a^{m+1-k}b^{k}\end{aligned}}}
考慮
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{\displaystyle (a+b)^{7}=(a+b)(a+b)(a+b)(a+b)(a+b)(a+b)(a+b)}
a
{\displaystyle a}
b
{\displaystyle b}
a
4
b
3
{\displaystyle a^{4}b^{3}}
(
7
4
)
{\displaystyle {\tbinom {7}{4}}}
(
7
4
)
{\displaystyle {\tbinom {7}{4}}}
a
4
b
3
{\displaystyle a^{4}b^{3}}
同理，
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n
=
(
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+
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)
(
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+
b
)
.
.
.
.
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+
b
)
(
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+
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{\displaystyle (a+b)^{n}=(a+b)(a+b)....(a+b)(a+b)}
n
{\displaystyle n}
n
{\displaystyle n}
k
{\displaystyle k}
a
{\displaystyle a}
(
n
−
k
)
{\displaystyle (n-k)}
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n
−
k
)
{\displaystyle (n-k)}
b
{\displaystyle b}
a
k
b
n
−
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{\displaystyle a^{k}b^{n-k}}
(
n
k
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{\displaystyle {\tbinom {n}{k}}}
(
n
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{\displaystyle {\tbinom {n}{k}}}
a
k
b
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−
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{\displaystyle a^{k}b^{n-k}}
考慮
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=
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{\displaystyle (a+b)^{7}=(a+b)(a+b)(a+b)(a+b)(a+b)(a+b)(a+b)}
a
{\displaystyle a}
b
{\displaystyle b}
a
{\displaystyle a}
b
{\displaystyle b}
a
a
a
a
b
b
b
{\displaystyle aaaabbb}
7
!
4
!
×
3
!
{\displaystyle {\frac {7!}{4!\times 3!}}}
(
7
4
)
{\displaystyle {\tbinom {7}{4}}}
同理，
(
a
+
b
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n
=
(
a
+
b
)
(
a
+
b
)
.
.
.
.
(
a
+
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(
a
+
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{\displaystyle (a+b)^{n}=(a+b)(a+b)....(a+b)(a+b)}
a
{\displaystyle a}
b
{\displaystyle b}
k
{\displaystyle k}
a
{\displaystyle a}
(
n
−
k
)
{\displaystyle (n-k)}
b
{\displaystyle b}
a
a
…
a
a
⏟
k
b
b
…
b
b
⏟
n
−
k
{\displaystyle \underbrace {aa\ldots aa} _{k}\underbrace {bb\ldots bb} _{n-k}}
n
!
k
!
×
(
n
−
k
)
!
{\displaystyle {\frac {n!}{k!\times (n-k)!}}}
(
n
k
)
{\displaystyle {\tbinom {n}{k}}}
通常二项式定理可以直接使用泰勒公式 进行证明. 下面的方法不使用泰勒公式
设
f
(
x
)
=
(
1
+
x
)
a
{\displaystyle f(x)=(1+x)^{a}}
g
(
x
)
=
∑
k
=
0
∞
(
a
k
)
x
k
{\displaystyle g(x)=\sum _{k=0}^{\infty }{a \choose k}x^{k}}
|
x
|
<
1
{\displaystyle |x|<1}
首先证明
g
(
x
)
{\displaystyle g(x)}
|
x
|
<
1
{\displaystyle |x|<1}
之后，易得
f
(
x
)
{\displaystyle f(x)}
(
1
+
x
)
f
′
(
x
)
=
a
f
(
x
)
{\displaystyle (1+x)f'(x)=af(x)}
再证明
g
(
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)
{\displaystyle g(x)}
g
(
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)
=
1
+
∑
k
=
1
∞
(
a
k
)
x
k
{\displaystyle g(x)=1+\sum _{k=1}^{\infty }{a \choose k}x^{k}}
g
′
(
x
)
=
∑
k
=
1
∞
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a
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x
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−
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=
∑
k
=
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∞
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a
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+
1
)
(
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+
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)
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k
=
∑
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=
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∞
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a
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)
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a
−
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)
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}}}
因为
(
a
k
+
1
)
(
k
+
1
)
=
(
a
)
(
a
−
1
)
⋯
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a
−
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+
1
)
(
a
−
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)
(
k
+
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)
!
(
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+
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)
=
(
a
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(
a
−
1
)
⋯
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a
−
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+
1
)
(
a
−
k
)
k
!
=
(
a
k
)
(
a
−
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)
{\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}}}
于是
(
1
+
x
)
g
′
(
x
)
=
g
′
(
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)
+
x
∑
k
=
1
∞
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k
x
k
−
1
=
∑
k
=
0
∞
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)
(
a
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)
x
k
+
∑
k
=
1
∞
(
a
k
)
k
x
k
=
∑
k
=
0
∞
(
a
k
)
(
a
−
k
)
x
k
+
∑
k
=
0
∞
(
a
k
)
k
x
k
=
∑
k
=
0
∞
(
a
k
)
x
k
(
a
−
k
+
k
)
=
a
∑
k
=
0
∞
(
a
k
)
x
k
=
a
⋅
g
(
x
)
{\displaystyle {\begin{aligned}(1+x)g'(x)&=g'(x)+x\sum _{k=1}^{\infty }{a \choose k}kx^{k-1}\\&=\sum _{k=0}^{\infty }{a \choose k}(a-k)x^{k}+\sum _{k=1}^{\infty }{a \choose k}kx^{k}\\&=\sum _{k=0}^{\infty }{a \choose k}(a-k)x^{k}+\sum _{k=0}^{\infty }{a \choose k}kx^{k}\\&=\sum _{k=0}^{\infty }{a \choose k}x^{k}(a-k+k)\\&=a\sum _{k=0}^{\infty }{a \choose k}x^{k}\\&=a\cdot g(x)\\\end{aligned}}}
因为
∑
k
=
0
∞
(
a
k
)
k
x
k
=
(
a
0
)
⋅
0
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+
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=
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∞
(
a
k
)
k
x
k
=
∑
k
=
1
∞
(
a
k
)
k
x
k
{\displaystyle \sum _{k=0}^{\infty }{a \choose k}kx^{k}={a \choose 0}\cdot 0x^{0}+\sum _{k=1}^{\infty }{a \choose k}kx^{k}=\sum _{k=1}^{\infty }{a \choose k}kx^{k}}
g
′
(
x
)
g
(
x
)
=
a
1
+
x
{\displaystyle {\frac {g'(x)}{g(x)}}={\frac {a}{1+x}}}
∵
f
′
(
x
)
f
(
x
)
=
a
1
+
x
{\displaystyle \because {\frac {f'(x)}{f(x)}}={\frac {a}{1+x}}}
∴
f
′
(
x
)
f
(
x
)
=
g
′
(
x
)
g
(
x
)
{\displaystyle \therefore {\frac {f'(x)}{f(x)}}={\frac {g'(x)}{g(x)}}}
g
′
(
x
)
f
(
x
)
=
f
′
(
x
)
g
(
x
)
{\displaystyle g'(x)f(x)=f'(x)g(x)}
根据除法定则 ，
d
d
x
(
g
(
x
)
f
(
x
)
)
=
g
′
(
x
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f
(
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)
−
f
′
(
x
)
g
(
x
)
(
f
(
x
)
)
2
=
0
{\displaystyle {\frac {d}{dx}}\left({\frac {g(x)}{f(x)}}\right)={\frac {g'(x)f(x)-f'(x)g(x)}{(f(x))^{2}}}=0}
根据拉格朗日中值定理 ，
g
(
x
)
f
(
x
)
{\displaystyle {\frac {g(x)}{f(x)}}}
常数函數 .
g
(
x
)
f
(
x
)
=
g
(
0
)
f
(
0
)
=
1
{\displaystyle {\frac {g(x)}{f(x)}}={\frac {g(0)}{f(0)}}=1}
f
(
x
)
=
g
(
x
)
{\displaystyle f(x)=g(x)}
牛顿以二项式定理作为基礎发明出了微积分 [ 10] 近似 、估计以及证明恒等式 等。
二项式定理给出的系数可以视为组合数
(
n
k
)
{\displaystyle {n \choose k}}
(1)证明
∑
k
=
0
n
(
n
k
)
2
=
(
2
n
n
)
{\displaystyle \sum _{k=0}^{n}{n \choose k}^{2}={2n \choose n}}
可以考虑恒等式
(
1
+
x
)
n
(
1
+
x
)
n
=
(
1
+
x
)
2
n
{\displaystyle (1+x)^{n}(1+x)^{n}=(1+x)^{2n}}
∑
i
=
0
n
∑
j
=
0
n
(
n
i
)
(
n
j
)
x
i
x
j
{\displaystyle \sum _{i=0}^{n}\sum _{j=0}^{n}{n \choose i}{n \choose j}x^{i}x^{j}}
∑
k
=
0
2
n
(
2
n
k
)
x
k
{\displaystyle \sum _{k=0}^{2n}{2n \choose k}x^{k}}
k
{\displaystyle k}
∑
i
=
0
k
(
n
i
)
(
n
k
−
i
)
=
(
2
n
k
)
{\displaystyle \sum _{i=0}^{k}{n \choose i}{n \choose k-i}={2n \choose k}}
k
=
n
{\displaystyle k=n}
(
n
i
)
=
(
n
n
−
i
)
{\displaystyle {n \choose i}={n \choose n-i}}
(2)證明
∑
k
=
0
n
(
n
k
)
=
2
n
{\displaystyle \sum _{k=0}^{n}{n \choose k}=2^{n}}
因為
(
x
+
y
)
n
=
∑
k
=
0
n
(
n
k
)
x
n
−
k
y
k
{\displaystyle (x+y)^{n}=\sum _{k=0}^{n}{n \choose k}x^{n-k}y^{k}}
令
x
=
y
=
1
{\displaystyle x=y=1}
(
1
+
1
)
n
=
2
n
=
∑
k
=
0
n
(
n
k
)
⋅
1
n
−
k
⋅
1
k
=
(
n
0
)
+
(
n
1
)
+
(
n
2
)
+
⋯
+
(
n
n
)
=
∑
k
=
0
n
(
n
k
)
{\displaystyle {\begin{aligned}(1+1)^{n}&=2^{n}=\sum _{k=0}^{n}{n \choose k}\cdot 1^{n-k}\cdot 1^{k}\\&={n \choose 0}+{n \choose 1}+{n \choose 2}+\cdots +{n \choose n}\\&=\sum _{k=0}^{n}{n \choose k}\end{aligned}}}
在复数 中，二项式定理可以與棣莫弗公式 結合，成為多倍角公式 [ 11]
cos
(
n
x
)
+
i
sin
(
n
x
)
=
(
cos
x
+
i
sin
x
)
n
.
{\displaystyle \cos \left(nx\right)+i\sin \left(nx\right)=\left(\cos x+i\sin x\right)^{n}.\,}
通過使用二项式定理，右邊的表達式可以擴展為
(
cos
x
+
i
sin
x
)
2
=
cos
2
x
+
2
i
cos
x
sin
x
−
sin
2
x
,
{\displaystyle \left(\cos x+i\sin x\right)^{2}=\cos ^{2}x+2i\cos x\sin x-\sin ^{2}x,}
由棣莫弗公式，实部与虚部对应，能夠得出
cos
(
2
x
)
=
cos
2
x
−
sin
2
x
and
sin
(
2
x
)
=
2
cos
x
sin
x
,
{\displaystyle \cos(2x)=\cos ^{2}x-\sin ^{2}x\quad {\text{and}}\quad \sin(2x)=2\cos x\sin x,}
即二倍角公式。同樣，因為
(
cos
x
+
i
sin
x
)
3
=
cos
3
x
+
3
i
cos
2
x
sin
x
−
3
cos
x
sin
2
x
−
i
sin
3
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,}
所以藉棣莫弗公式，能夠得出
cos
(
3
x
)
=
cos
3
x
−
3
cos
x
sin
2
x
and
sin
(
3
x
)
=
3
cos
2
x
sin
x
−
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.}
整體而言，多倍角恒等式可以寫作
cos
(
n
x
)
=
∑
k
even
(
−
1
)
k
2
(
n
k
)
cos
n
−
k
x
sin
k
x
{\displaystyle \cos(nx)=\sum _{k{\text{ even}}}(-1)^{\frac {k}{2}}{n \choose k}\cos ^{n-k}x\sin ^{k}x}
和
sin
(
n
x
)
=
∑
k
odd
(
−
1
)
k
−
1
2
(
n
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.}
數學常數 e 的定義爲下列極限 值：[ 12]
e
=
lim
n
→
∞
(
1
+
1
n
)
n
.
{\displaystyle e=\lim _{n\to \infty }\left(1+{\frac {1}{n}}\right)^{n}.}
使用二项式定理能得出
(
1
+
1
n
)
n
=
1
+
(
n
1
)
1
n
+
(
n
2
)
1
n
2
+
(
n
3
)
1
n
3
+
⋯
+
(
n
n
)
1
n
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}}}.}
第
k
{\displaystyle k}
(
n
k
)
1
n
k
=
1
k
!
⋅
n
(
n
−
1
)
(
n
−
2
)
⋯
(
n
−
k
+
1
)
n
k
{\displaystyle {n \choose k}{\frac {1}{n^{k}}}\;=\;{\frac {1}{k!}}\cdot {\frac {n(n-1)(n-2)\cdots (n-k+1)}{n^{k}}}}
因為
n
→
∞
{\displaystyle n\to \infty }
lim
n
→
∞
(
n
k
)
1
n
k
=
1
k
!
.
{\displaystyle \lim _{n\to \infty }{n \choose k}{\frac {1}{n^{k}}}={\frac {1}{k!}}.}
這表明
e
{\displaystyle e}
[ 13] [ 14]
e
=
∑
k
=
0
∞
1
k
!
=
1
0
!
+
1
1
!
+
1
2
!
+
1
3
!
+
⋯
.
{\displaystyle e=\sum _{k=0}^{\infty }{\frac {1}{k!}}={\frac {1}{0!}}+{\frac {1}{1!}}+{\frac {1}{2!}}+{\frac {1}{3!}}+\cdots .}
该定理可以推广到对任意实数次幂的展开，即所谓的牛顿广义二项式定理 ：
(
x
+
y
)
α
=
∑
k
=
0
∞
(
α
k
)
x
α
−
k
y
k
{\displaystyle (x+y)^{\alpha }=\sum _{k=0}^{\infty }{\alpha \choose k}x^{\alpha -k}y^{k}}
(
α
k
)
=
α
(
α
−
1
)
.
.
.
(
α
−
k
+
1
)
k
!
=
(
α
)
k
k
!
{\displaystyle {\alpha \choose k}={\frac {\alpha (\alpha -1)...(\alpha -k+1)}{k!}}={\frac {(\alpha )_{k}}{k!}}}
对于多元形式的多项式展开，可以看做二项式定理的推广：[ 15] [ 16]
(
x
1
+
x
2
+
.
.
.
+
x
n
)
k
=
∑
α
1
+
α
2
+
.
.
.
+
α
n
=
k
k
!
α
1
!
.
.
.
α
n
!
x
1
α
1
.
.
.
x
n
α
n
{\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}}}
证明：
数学归纳法 。对元数
n
{\displaystyle n}
n
=
2
{\displaystyle n=2}
n
−
1
{\displaystyle n-1}
(
x
1
+
x
2
+
.
.
.
+
x
n
)
k
=
(
(
x
1
+
x
2
+
.
.
.
+
x
n
−
1
)
+
x
n
)
k
=
∑
α
n
=
0
k
k
!
α
n
!
(
k
−
α
n
)
!
(
x
1
+
x
2
+
.
.
.
+
x
n
−
1
)
k
−
α
n
x
n
α
n
=
∑
α
n
=
0
k
k
!
α
n
!
(
k
−
α
n
)
!
∑
α
1
+
α
2
+
.
.
.
+
α
n
−
1
=
k
−
α
n
(
k
−
α
n
)
!
α
1
!
.
.
.
α
n
−
1
!
x
1
α
1
.
.
.
x
n
−
1
α
n
−
1
x
n
α
n
=
∑
α
1
+
α
2
+
.
.
.
+
α
n
=
k
k
!
α
1
!
.
.
.
α
n
!
x
1
α
1
.
.
.
x
n
α
n
證 畢
{\displaystyle {\begin{aligned}\left(x_{1}+x_{2}+...+x_{n}\right)^{k}&=((x_{1}+x_{2}+...+x_{n-1})+x_{n})^{k}\\&=\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}}\\&=\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}}\\&=\sum _{\alpha _{1}+\alpha _{2}+...+\alpha _{n}=k}{\frac {k!}{\alpha _{1}!...\alpha _{n}!}}x_{1}^{\alpha _{1}}...x_{n}^{\alpha _{n}}{\text{證 畢 }}\\\end{aligned}}}
^ Binomial Expansions - leeds uk (PDF) . [2015-04-12 ] . （原始内容存档 (PDF) 于2020-09-19）. ^ Roman, Steven "The Umbral Calculus", Dover Publications, 2005, ISBN 0-486-44129-3
^ 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.
^ Weisstein, Eric W. (编). Binomial Theorem . at MathWorld Wolfram Research, Inc. [2015-04-11 ] . （原始内容存档 于2020-11-14） （英语） . ^ The Story of the Binomial Theorem by J. L. Coolidge , The American Mathematical Monthly 56 :3 (1949), pp. 147–157^ 6.0 6.1 約翰·J·奧康納; 埃德蒙·F·羅伯遜 , Abu Bekr ibn Muhammad ibn al-Husayn Al-Karaji , MacTutor数学史档案 （英语） ^ Landau, James A. Historia Matematica Mailing List Archive: Re: [HM] Pascal's Triangle . Archives of Historia Matematica. 1999-05-08 [2007-04-13 ] . （原始内容 (mailing list email) 存档于2021-02-24）. ^ Bourbaki: History of mathematics
^ Barth, Nils R. Computing Cavalieri's Quadrature Formula by a Symmetry of the n -Cube . The American Mathematical Monthly. 2004, 111 (9): 811–813. ISSN 0002-9890 JSTOR 4145193 doi:10.2307/4145193 author's copy , further remarks and resources ^ 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 ^ Weisstein, Eric W. (编). Multiple-Angle Formulas . at MathWorld Wolfram Research, Inc. [2015-04-12 ] . （原始内容存档 于2020-11-11） （英语） . ^ The Constant e - NDE/NDT Resource Center . [2015-04-12 ] . （原始内容存档 于2020-11-11）. ^ Series - NTEC (PDF) . [2015-04-12 ] . （原始内容存档 (PDF) 于2020-09-23）. ^ Encyclopedic Dictionary of Mathematics 142.D^ 多項式定理的新證明及其展開 - 佛山科学技术学院信息科学与数学系 . [2015-04-11 ] . （原始内容存档 于2017-04-13）. ^ Hazewinkel, Michiel (编), Multinomial coefficient , 数学百科全书 , Springer , 2001, ISBN 978-1-55608-010-4
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 ,
Graham, Ronald; Knuth, Donald; Patashnik, Oren. (5) Binomial Coefficients. Concrete Mathematics 2nd. Addison Wesley. 1994: 153 –256. ISBN 0-201-55802-5OCLC 17649857
![{\displaystyle {\begin{aligned}(a+b)^{m+1}&=a(a+b)^{m}+b(a+b)^{m}\\&=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}\\&=\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}\ \ \ \ \ {\text{( 將 }}a{\text{, }}b{\text{) }}\\&=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}\ \ \ \ \ {\text{ 取 出 }}k=0{\text{ 的 項 }}\\&=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}\ \ \ \ \ {\text{設 }}j=k-1\\&=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}\ \ \ \ \ {\text{取 出 }}k=m+1{\text{項 }}\\&=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}\ \ \ \ \ {\text{兩 者 加 起 }}\\&=a^{m+1}+b^{m+1}+\sum _{k=1}^{m}{m+1 \choose k}a^{m+1-k}b^{k}\ \ \ \ \ {\text{套 用 帕 斯 卡 法 則 }}\\&=\sum _{k=0}^{m+1}{m+1 \choose k}a^{m+1-k}b^{k}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/06677b47f9743554492cd5a4279abc3d01d75aff)
, doi:10.1007/s10699-012-9285-8
