首先給出 Binomial Theorem 的陳述
Theorem:
令 $a,b \in \mathbb{R}$ 且 $n \in \mathbb{N}$ 則
\[{(a + b)^n} = \mathop \sum \limits_{k = 0}^n \left( \begin{array}{l}
n\\
k
\end{array} \right){a^k}{b^{n - k}}\]
在證明 Binomial Theorem之前，我們會需要以下結果來幫助我們，
FACT: 給定 $n,k \in \mathbb{N}$ 且 $n \ge k$，則
\[\left( \begin{array}{l}
n + 1\\
k
\end{array} \right) = \left( \begin{array}{l}
n\\
k
\end{array} \right) + \left( \begin{array}{l}
n\\
k - 1
\end{array} \right)\]
Proof: 觀察左式，由定義可知
\[\left( \begin{array}{l}
n + 1\\
k
\end{array} \right): = \frac{{\left( {n + 1} \right)!}}{{k!\left( {n + 1 - k} \right)!}}\]
故如果我們可以證明右式等同於左式，則證明完畢。現在觀察右式
\[\begin{array}{l}
\left( \begin{array}{l}
n\\
k
\end{array} \right) + \left( \begin{array}{l}
n\\
k - 1
\end{array} \right) = \frac{{n!}}{{k!\left( {n - k} \right)!}} + \frac{{n!}}{{\left( {k - 1} \right)!\left( {n - \left( {k - 1} \right)} \right)!}}\\
\begin{array}{*{20}{c}}
{}&{}&{}&{}&{}
\end{array} = \frac{{n!\left( {n - k + 1} \right)}}{{k!\left( {n - k + 1} \right)\left( {n - k} \right)!}} + \frac{{n!k}}{{k\left( {k - 1} \right)!\left( {n + 1 - k} \right)!}}\\
\begin{array}{*{20}{c}}
{}&{}&{}&{}&{}
\end{array} = \frac{{n!\left( {n - k + 1} \right) + n!k}}{{k!\left( {n + 1 - k} \right)!}}\\
\begin{array}{*{20}{c}}
{}&{}&{}&{}&{}
\end{array} = \frac{{n!\left( {n + 1} \right)}}{{k!\left( {n + 1 - k} \right)!}} = \frac{{\left( {n + 1} \right)!}}{{k!\left( {n + 1 - k} \right)!}}
\end{array}\]注意到最後的等式與原本左式相等，故證明完畢。$\square$
現在我們可以開始證明 Binomial Theorem
Proof: (Binomial Theorem)
給定 $a,b \in \mathbb{R}$ 且 $n \in \mathbb{N}$ 利用數學歸納法，先觀察 $n=1$ 則
\[\begin{array}{l}
{(a + b)^1} = \sum\limits_{k = 0}^1 {\left( {\begin{array}{*{20}{l}}
1\\
k
\end{array}} \right)} {a^k}{b^{1 - k}}\\
\begin{array}{*{20}{c}}
{}&{}&{}
\end{array} = \left( {\begin{array}{*{20}{l}}
1\\
0
\end{array}} \right){a^0}{b^{1 - 0}} + \left( {\begin{array}{*{20}{l}}
1\\
1
\end{array}} \right){a^1}{b^{1 - 1}}\\
\begin{array}{*{20}{c}}
{}&{}&{}
\end{array} = {b^1} + {a^1}
\end{array}\]故成立。現在利用歸納法假設對 Binomial Theorem 對 $n$ 成立，我們要證明 $n+1$ 也成立：亦即假設 ${(a + b)^n} = \mathop \sum \limits_{k = 0}^n \left( \begin{array}{l}
n\\
k
\end{array} \right){a^k}{b^{n - k}}$成立，我們要證明
\[{(a + b)^{n+1}} = \mathop \sum \limits_{k = 0}^{n+1} \left( \begin{array}{l}
n+1\\
k
\end{array} \right){a^k}{b^{n+1 - k}} \;\;\;\;\; (*)
\]現在觀察
\[\begin{array}{l}
{\left( {a + b} \right)^{n + 1}} = {\left( {a + b} \right)^n}\left( {a + b} \right)\\
= \sum\limits_{k = 0}^n {\left( {\begin{array}{*{20}{l}}
n\\
k
\end{array}} \right)} {a^k}{b^{n - k}}\left( {a + b} \right)\\
= \sum\limits_{k = 0}^n {\left( {\begin{array}{*{20}{l}}
n\\
k
\end{array}} \right)} {a^{k + 1}}{b^{n - k}} + \sum\limits_{k = 0}^n {\left( {\begin{array}{*{20}{l}}
n\\
k
\end{array}} \right)} {a^k}{b^{n - k + 1}}
\end{array}\]為了使上述兩項和可以合成，我們使用變數變換，對第一項和令 $j=k+1$ 且對第二項和令 $j=k$ 則我們可得
\[\begin{array}{*{20}{l}}
{{{\left( {a + b} \right)}^{n + 1}} = \sum\limits_{k = 0}^n {\left( {\begin{array}{*{20}{l}}
n\\
k
\end{array}} \right)} {a^{k + 1}}{b^{n - k}} + \sum\limits_{k = 0}^n {\left( {\begin{array}{*{20}{l}}
n\\
k
\end{array}} \right)} {a^k}{b^{n - k + 1}}}\\
{\begin{array}{*{20}{c}}
{}&{}&{}&{}
\end{array} = \underbrace {\sum\limits_{j = 1}^{n + 1} {\left( {\begin{array}{*{20}{l}}
n\\
{j - 1}
\end{array}} \right)} {a^j}{b^{n - \left( {j - 1} \right)}}}_{term1} + \underbrace {\sum\limits_{j = 0}^n {\left( {\begin{array}{*{20}{l}}
n\\
j
\end{array}} \right)} {a^j}{b^{n - j + 1}}}_{term2}}\\
\begin{array}{l}
\begin{array}{*{20}{c}}
{}&{}&{}&{}
\end{array} = \underbrace {\left( {\begin{array}{*{20}{l}}
n\\
n
\end{array}} \right){a^{\left( {n + 1} \right)}}{b^{n - \left( {n + 1} \right) + 1}} + \sum\limits_{j = 1}^n {\left( {\begin{array}{*{20}{l}}
n\\
{j - 1}
\end{array}} \right)} {a^j}{b^{n - j + 1}}}_{ = term1}\\
\begin{array}{*{20}{c}}
{}&{}&{}&{}&{}&{}&{}&{}
\end{array} + \underbrace {\sum\limits_{j = 1}^n {\left( {\begin{array}{*{20}{l}}
n\\
j
\end{array}} \right)} {a^j}{b^{n - j + 1}} + \left( {\begin{array}{*{20}{l}}
n\\
0
\end{array}} \right){a^0}{b^{n - 0 + 1}}}_{ = term2}
\end{array}\\
{\begin{array}{*{20}{c}}
{}&{}&{}&{}
\end{array} = {a^{\left( {n + 1} \right)}} + \sum\limits_{j = 1}^n {\left[ {\left( {\begin{array}{*{20}{l}}
n\\
{j - 1}
\end{array}} \right) + \left( {\begin{array}{*{20}{l}}
n\\
j
\end{array}} \right)} \right]} {a^j}{b^{n - j + 1}} + {b^{n + 1}}}
\end{array}\]現在對中間項的和，使用前述的 FACT 可得
\[\begin{array}{l}
{\left( {a + b} \right)^{n + 1}} = {a^{\left( {n + 1} \right)}}{b^{n - \left( {n + 1} \right) + 1}} + \sum\limits_{j = 1}^n {\left[ {\left( {\begin{array}{*{20}{l}}
n\\
{j - 1}
\end{array}} \right) + \left( {\begin{array}{*{20}{l}}
n\\
j
\end{array}} \right)} \right]} {a^j}{b^{n - j + 1}} + {a^0}{b^{n - 0 + 1}}\\
\begin{array}{*{20}{c}}
{}&{}&{}&{}
\end{array} = {a^{\left( {n + 1} \right)}}{b^0} + \sum\limits_{j = 1}^n {\left( {\begin{array}{*{20}{l}}
{n + 1}\\
j
\end{array}} \right)} {a^j}{b^{n + 1 - j}} + {a^0}{b^{n + 1}}\\
\begin{array}{*{20}{c}}
{}&{}&{}&{}
\end{array} = \left( {\begin{array}{*{20}{l}}
{n + 1}\\
{n + 1}
\end{array}} \right){a^{\left( {n + 1} \right)}}{b^0} + \sum\limits_{j = 1}^n {\left( {\begin{array}{*{20}{l}}
{n + 1}\\
j
\end{array}} \right)} {a^j}{b^{n + 1 - j}} + \left( {\begin{array}{*{20}{l}}
{n + 1}\\
0
\end{array}} \right){a^0}{b^{n + 1}}\\
\begin{array}{*{20}{c}}
{}&{}&{}&{}
\end{array} = \sum\limits_{j = 0}^{n + 1} {\left( {\begin{array}{*{20}{l}}
{n + 1}\\
j
\end{array}} \right)} {a^j}{b^{n + 1 - j}}
\end{array}\]上式滿足 $(*)$ 故得證。 $\square$
以下我們看幾個例子：
Example 1:
試求\[\mathop \sum \limits_{i = 0}^n \left( \begin{array}{l}
n\\
i
\end{array} \right) = ?\]
Proof: 利用 Binomial Theorem，令 $a=b=1$ 可立刻得到
\[\begin{array}{l}
{(a + b)^n} = \sum\limits_{k = 0}^n {\left( {\begin{array}{*{20}{l}}
n\\
k
\end{array}} \right)} {a^k}{b^{n - k}}\\
\Rightarrow \sum\limits_{k = 0}^n {\left( {\begin{array}{*{20}{l}}
n\\
k
\end{array}} \right)} {1^k}{1^{n - k}} = {(1 + 1)^n} = {2^n}
\end{array}\]
Example 2:
試求
\[\mathop \sum \limits_{i = 0}^n {\left( { - 1} \right)^i}\left( \begin{array}{l}
n\\
i
\end{array} \right) = ?\]
Proof: 利用 Binomial Theorem，令 $a=-1$ 與 $b=1$ 可立刻得到
\[\begin{array}{l}
{(a + b)^n} = \sum\limits_{k = 0}^n {\left( {\begin{array}{*{20}{l}}
n\\
k
\end{array}} \right)} {a^k}{b^{n - k}}\\
\Rightarrow \sum\limits_{k = 0}^n {\left( {\begin{array}{*{20}{l}}
n\\
k
\end{array}} \right)} {\left( { - 1} \right)^k}{1^{n - k}} = {( - 1 + 1)^n} = {0^n} = 0
\end{array}\]
Theorem:
令 $a,b \in \mathbb{R}$ 且 $n \in \mathbb{N}$ 則
\[{(a + b)^n} = \mathop \sum \limits_{k = 0}^n \left( \begin{array}{l}
n\\
k
\end{array} \right){a^k}{b^{n - k}}\]
在證明 Binomial Theorem之前，我們會需要以下結果來幫助我們，
FACT: 給定 $n,k \in \mathbb{N}$ 且 $n \ge k$，則
\[\left( \begin{array}{l}
n + 1\\
k
\end{array} \right) = \left( \begin{array}{l}
n\\
k
\end{array} \right) + \left( \begin{array}{l}
n\\
k - 1
\end{array} \right)\]
Proof: 觀察左式，由定義可知
\[\left( \begin{array}{l}
n + 1\\
k
\end{array} \right): = \frac{{\left( {n + 1} \right)!}}{{k!\left( {n + 1 - k} \right)!}}\]
故如果我們可以證明右式等同於左式，則證明完畢。現在觀察右式
\[\begin{array}{l}
\left( \begin{array}{l}
n\\
k
\end{array} \right) + \left( \begin{array}{l}
n\\
k - 1
\end{array} \right) = \frac{{n!}}{{k!\left( {n - k} \right)!}} + \frac{{n!}}{{\left( {k - 1} \right)!\left( {n - \left( {k - 1} \right)} \right)!}}\\
\begin{array}{*{20}{c}}
{}&{}&{}&{}&{}
\end{array} = \frac{{n!\left( {n - k + 1} \right)}}{{k!\left( {n - k + 1} \right)\left( {n - k} \right)!}} + \frac{{n!k}}{{k\left( {k - 1} \right)!\left( {n + 1 - k} \right)!}}\\
\begin{array}{*{20}{c}}
{}&{}&{}&{}&{}
\end{array} = \frac{{n!\left( {n - k + 1} \right) + n!k}}{{k!\left( {n + 1 - k} \right)!}}\\
\begin{array}{*{20}{c}}
{}&{}&{}&{}&{}
\end{array} = \frac{{n!\left( {n + 1} \right)}}{{k!\left( {n + 1 - k} \right)!}} = \frac{{\left( {n + 1} \right)!}}{{k!\left( {n + 1 - k} \right)!}}
\end{array}\]注意到最後的等式與原本左式相等，故證明完畢。$\square$
現在我們可以開始證明 Binomial Theorem
Proof: (Binomial Theorem)
給定 $a,b \in \mathbb{R}$ 且 $n \in \mathbb{N}$ 利用數學歸納法，先觀察 $n=1$ 則
\[\begin{array}{l}
{(a + b)^1} = \sum\limits_{k = 0}^1 {\left( {\begin{array}{*{20}{l}}
1\\
k
\end{array}} \right)} {a^k}{b^{1 - k}}\\
\begin{array}{*{20}{c}}
{}&{}&{}
\end{array} = \left( {\begin{array}{*{20}{l}}
1\\
0
\end{array}} \right){a^0}{b^{1 - 0}} + \left( {\begin{array}{*{20}{l}}
1\\
1
\end{array}} \right){a^1}{b^{1 - 1}}\\
\begin{array}{*{20}{c}}
{}&{}&{}
\end{array} = {b^1} + {a^1}
\end{array}\]故成立。現在利用歸納法假設對 Binomial Theorem 對 $n$ 成立，我們要證明 $n+1$ 也成立：亦即假設 ${(a + b)^n} = \mathop \sum \limits_{k = 0}^n \left( \begin{array}{l}
n\\
k
\end{array} \right){a^k}{b^{n - k}}$成立，我們要證明
\[{(a + b)^{n+1}} = \mathop \sum \limits_{k = 0}^{n+1} \left( \begin{array}{l}
n+1\\
k
\end{array} \right){a^k}{b^{n+1 - k}} \;\;\;\;\; (*)
\]現在觀察
\[\begin{array}{l}
{\left( {a + b} \right)^{n + 1}} = {\left( {a + b} \right)^n}\left( {a + b} \right)\\
= \sum\limits_{k = 0}^n {\left( {\begin{array}{*{20}{l}}
n\\
k
\end{array}} \right)} {a^k}{b^{n - k}}\left( {a + b} \right)\\
= \sum\limits_{k = 0}^n {\left( {\begin{array}{*{20}{l}}
n\\
k
\end{array}} \right)} {a^{k + 1}}{b^{n - k}} + \sum\limits_{k = 0}^n {\left( {\begin{array}{*{20}{l}}
n\\
k
\end{array}} \right)} {a^k}{b^{n - k + 1}}
\end{array}\]為了使上述兩項和可以合成，我們使用變數變換，對第一項和令 $j=k+1$ 且對第二項和令 $j=k$ 則我們可得
\[\begin{array}{*{20}{l}}
{{{\left( {a + b} \right)}^{n + 1}} = \sum\limits_{k = 0}^n {\left( {\begin{array}{*{20}{l}}
n\\
k
\end{array}} \right)} {a^{k + 1}}{b^{n - k}} + \sum\limits_{k = 0}^n {\left( {\begin{array}{*{20}{l}}
n\\
k
\end{array}} \right)} {a^k}{b^{n - k + 1}}}\\
{\begin{array}{*{20}{c}}
{}&{}&{}&{}
\end{array} = \underbrace {\sum\limits_{j = 1}^{n + 1} {\left( {\begin{array}{*{20}{l}}
n\\
{j - 1}
\end{array}} \right)} {a^j}{b^{n - \left( {j - 1} \right)}}}_{term1} + \underbrace {\sum\limits_{j = 0}^n {\left( {\begin{array}{*{20}{l}}
n\\
j
\end{array}} \right)} {a^j}{b^{n - j + 1}}}_{term2}}\\
\begin{array}{l}
\begin{array}{*{20}{c}}
{}&{}&{}&{}
\end{array} = \underbrace {\left( {\begin{array}{*{20}{l}}
n\\
n
\end{array}} \right){a^{\left( {n + 1} \right)}}{b^{n - \left( {n + 1} \right) + 1}} + \sum\limits_{j = 1}^n {\left( {\begin{array}{*{20}{l}}
n\\
{j - 1}
\end{array}} \right)} {a^j}{b^{n - j + 1}}}_{ = term1}\\
\begin{array}{*{20}{c}}
{}&{}&{}&{}&{}&{}&{}&{}
\end{array} + \underbrace {\sum\limits_{j = 1}^n {\left( {\begin{array}{*{20}{l}}
n\\
j
\end{array}} \right)} {a^j}{b^{n - j + 1}} + \left( {\begin{array}{*{20}{l}}
n\\
0
\end{array}} \right){a^0}{b^{n - 0 + 1}}}_{ = term2}
\end{array}\\
{\begin{array}{*{20}{c}}
{}&{}&{}&{}
\end{array} = {a^{\left( {n + 1} \right)}} + \sum\limits_{j = 1}^n {\left[ {\left( {\begin{array}{*{20}{l}}
n\\
{j - 1}
\end{array}} \right) + \left( {\begin{array}{*{20}{l}}
n\\
j
\end{array}} \right)} \right]} {a^j}{b^{n - j + 1}} + {b^{n + 1}}}
\end{array}\]現在對中間項的和，使用前述的 FACT 可得
\[\begin{array}{l}
{\left( {a + b} \right)^{n + 1}} = {a^{\left( {n + 1} \right)}}{b^{n - \left( {n + 1} \right) + 1}} + \sum\limits_{j = 1}^n {\left[ {\left( {\begin{array}{*{20}{l}}
n\\
{j - 1}
\end{array}} \right) + \left( {\begin{array}{*{20}{l}}
n\\
j
\end{array}} \right)} \right]} {a^j}{b^{n - j + 1}} + {a^0}{b^{n - 0 + 1}}\\
\begin{array}{*{20}{c}}
{}&{}&{}&{}
\end{array} = {a^{\left( {n + 1} \right)}}{b^0} + \sum\limits_{j = 1}^n {\left( {\begin{array}{*{20}{l}}
{n + 1}\\
j
\end{array}} \right)} {a^j}{b^{n + 1 - j}} + {a^0}{b^{n + 1}}\\
\begin{array}{*{20}{c}}
{}&{}&{}&{}
\end{array} = \left( {\begin{array}{*{20}{l}}
{n + 1}\\
{n + 1}
\end{array}} \right){a^{\left( {n + 1} \right)}}{b^0} + \sum\limits_{j = 1}^n {\left( {\begin{array}{*{20}{l}}
{n + 1}\\
j
\end{array}} \right)} {a^j}{b^{n + 1 - j}} + \left( {\begin{array}{*{20}{l}}
{n + 1}\\
0
\end{array}} \right){a^0}{b^{n + 1}}\\
\begin{array}{*{20}{c}}
{}&{}&{}&{}
\end{array} = \sum\limits_{j = 0}^{n + 1} {\left( {\begin{array}{*{20}{l}}
{n + 1}\\
j
\end{array}} \right)} {a^j}{b^{n + 1 - j}}
\end{array}\]上式滿足 $(*)$ 故得證。 $\square$
以下我們看幾個例子：
Example 1:
試求\[\mathop \sum \limits_{i = 0}^n \left( \begin{array}{l}
n\\
i
\end{array} \right) = ?\]
Proof: 利用 Binomial Theorem，令 $a=b=1$ 可立刻得到
\[\begin{array}{l}
{(a + b)^n} = \sum\limits_{k = 0}^n {\left( {\begin{array}{*{20}{l}}
n\\
k
\end{array}} \right)} {a^k}{b^{n - k}}\\
\Rightarrow \sum\limits_{k = 0}^n {\left( {\begin{array}{*{20}{l}}
n\\
k
\end{array}} \right)} {1^k}{1^{n - k}} = {(1 + 1)^n} = {2^n}
\end{array}\]
Example 2:
試求
\[\mathop \sum \limits_{i = 0}^n {\left( { - 1} \right)^i}\left( \begin{array}{l}
n\\
i
\end{array} \right) = ?\]
Proof: 利用 Binomial Theorem，令 $a=-1$ 與 $b=1$ 可立刻得到
\[\begin{array}{l}
{(a + b)^n} = \sum\limits_{k = 0}^n {\left( {\begin{array}{*{20}{l}}
n\\
k
\end{array}} \right)} {a^k}{b^{n - k}}\\
\Rightarrow \sum\limits_{k = 0}^n {\left( {\begin{array}{*{20}{l}}
n\\
k
\end{array}} \right)} {\left( { - 1} \right)^k}{1^{n - k}} = {( - 1 + 1)^n} = {0^n} = 0
\end{array}\]
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