『書不盡言，言不盡意』為什麼呢？因為知識能積累、觸類可旁通、依樣畫葫蘆……，自然而然之事乎！然而論善用『形式操作符號』的人，豈非奇怪焉？設無『生之累，病之苦』，將為之若何哉？？此真

二項式定理

在初等代數中，二項式定理（英語：Binomial theorem）描述了二項式的冪的代數展開。根據該定理，可以將兩個數之和的整數次冪諸如(x + y)ⁿ 展開為類似 ax^by^c 項之和的恆等式，其中b、c均為非負整數且b + c = n。係數a是依賴於n和b的正整數。當某項的指數為0時，通常略去不寫。例如：^[1]

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

ax^by^c 中的係數a被稱為二項式係數，記作 ${\tbinom nb}$ 或 ${\tbinom nc}$ （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]。

二項式係數出現在楊輝三角（帕斯卡三角）中。除邊緣的數字外，其他每一個數都為其上方兩數之和。

所以歷史上『一再發現』

歷史

二項式係數的三角形排列通常被認為是法國數學家布萊茲·帕斯卡的貢獻，他在17世紀描述了這一現象^[3]。但早在他之前，就曾有數學家進行類似的研究。例如，古希臘數學家歐幾里得於公元前4世紀提到了指數為2的情況^[4]^[5]。公元前三世紀，印度數學家青目探討了更高階的情況。「帕斯卡三角形」的雛形於10世紀由印度數學家大力羅摩發現。在同一時期，波斯數學家卡拉吉^[6]和數學家兼詩人歐瑪爾·海亞姆得到了更為普遍的二項式定理的形式。13世紀，中國數學家楊輝也得到了類似的結果^[7]。卡拉吉用數學歸納法的原始形式給出了二項式定理和帕斯卡三角形的有關證明^[6]。艾薩克·牛頓勳爵將二項式定理的係數推廣到有理數^[8]。

！！？？甚至直至今日，難得人人皆知咦？？！！

莫非那『二項式定理』

定理的陳述

根據此定理，可以將 x + y 的任意次冪展開成和的形式

(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},

其中每個 ${\tbinom nk}$ 為一個稱作二項式係數的特定正整數，其等於 ${\frac {n!}{k!(n-k)!}}$ 。這個公式也稱二項式公式或二項恆等式。使用求和符號，可以把它寫作

(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}}.

後面的表達式只是將根據 x 與 y 的對稱性得出的，通過比較發現公式中的二項式係數也是對稱的。二項式定理的一個變形是用 1 來代換 y 得到的，所以它只涉及一個變量。在這種形式中，公式寫作

(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},

或者等價地

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

果不清晰明白，空有『多重詮釋』

Geometric explanation

For positive values of a and b, the binomial theorem with n = 2 is the geometrically evident fact that a square of side a + b can be cut into a square of side a, a square of side b, and two rectangles with sides a and b. With n = 3, the theorem states that a cube of side a + b can be cut into a cube of side a, a cube of side b, three a×a×b rectangular boxes, and three a×b×b rectangular boxes.

In calculus, this picture also gives a geometric proof of the derivative $(x^{n})'=nx^{n-1}:$ ^[9] if one sets $a=x$ and $b=\Delta x,$ interpreting b as an infinitesimal change in a, then this picture shows the infinitesimal change in the volume of an n-dimensional hypercube, $(x+\Delta x)^{n},$ where the coefficient of the linear term (in $\Delta x$ ) is $nx^{n-1},$ the area of the n faces, each of dimension $(n-1):$

(x+\Delta x)^{n}=x^{n}+nx^{n-1}\Delta x+{\tbinom {n}{2}}x^{n-2}(\Delta x)^{2}+\cdots .

Substituting this into the definition of the derivative via a difference quotient and taking limits means that the higher order terms, $(\Delta x)^{2}$ and higher, become negligible, and yields the formula $(x^{n})'=nx^{n-1},$ interpreted as

“the infinitesimal rate of change in volume of an n-cube as side length varies is the area of n of its

(n-1)

-dimensional faces”.

If one integrates this picture, which corresponds to applying the fundamental theorem of calculus, one obtains Cavalieri’s quadrature formula, the integral $\textstyle {\int x^{n-1}\,dx={\tfrac {1}{n}}x^{n}}$ – see proof of Cavalieri’s quadrature formula for details.^[9]

Visualisation of binomial expansion up to the 4th power

嗎？卻不知早播下『排列組合』之種子◎

Proofs

Combinatorial proof

Example

The coefficient of xy² in

{\begin{aligned}(x+y)^{3}&=(x+y)(x+y)(x+y)\\&=xxx+xxy+xyx+{\underline {xyy}}+yxx+{\underline {yxy}}+{\underline {yyx}}+yyy\\&=x^{3}+3x^{2}y+{\underline {3xy^{2}}}+y^{3}.\end{aligned}}

equals ${\tbinom {3}{2}}=3$ because there are three x,y strings of length 3 with exactly two y’s, namely,

xyy,\;yxy,\;yyx,

corresponding to the three 2-element subsets of { 1, 2, 3 }, namely,

\{2,3\},\;\{1,3\},\;\{1,2\},

where each subset specifies the positions of the y in a corresponding string.

General case

Expanding (x + y)ⁿ yields the sum of the 2ⁿ products of the form e₁e₂ … e_n where each e_i is x or y. Rearranging factors shows that each product equals x^n−ky^k for some k between 0 and n. For a given k, the following are proved equal in succession:

the number of copies of x^n − ky^k in the expansion
the number of n-character x,y strings having y in exactly k positions
the number of k-element subsets of { 1, 2, …, n}
${n \choose k}$ (this is either by definition, or by a short combinatorial argument if one is defining ${n \choose k}$ as ${\frac {n!}{k!(n-k)!}}$ ).

This proves the binomial theorem.

Inductive proof

Induction yields another proof of the binomial theorem. When n = 0, both sides equal 1, since x⁰ = 1 and ${\tbinom {0}{0}}=1$ . Now suppose that the equality holds for a given n; we will prove it for n + 1. For j, k ≥ 0, let [ƒ(x, y)]_j,k denote the coefficient of x^jy^k in the polynomial ƒ(x, y). By the inductive hypothesis, (x + y)ⁿ is a polynomial in x and y such that [(x + y)ⁿ]_j,k is ${\tbinom {n}{k}}$ if j + k = n, and 0 otherwise. The identity $(x+y)^{n+1}=x(x+y)^{n}+y(x+y)^{n}$