The Binomial Theorem

Consider the expansion of the binomial $(1 + x)^n$ for $n \in \{ 0, 1, 2, ... \}$. When $n = 0$ we have that:

(1)

\begin{align} \quad (1 + x)^0 = 1 \end{align}

When $n = 1$, $n = 2$, and $n = 3$ we get:

(2)

\begin{equation} (1 + x)^1 = 1 + x \end{equation}

(3)

\begin{align} \quad (1 + x)^2 = 1 + 2x + x^2 \end{align}

(4)

\begin{align} \quad (1 + x)^3 = 1 + 3x + 3x^2 + x^3 \end{align}

Notice that if we list the terms of the expansion of $(1 + x)^n$ in ascending order then the coefficients of these terms match the numbers in each row $n$ of Pascal's Triangle. More generally, this property is also apparent in the expansion of the binomial $(x + y)^n$. For $n = 0$, $n = 1$, $n = 2$, and $n = 3$ we have that:

(5)

\begin{align} \quad (x + y)^0 = 1 \end{align}

(6)

\begin{align} \quad (x + y)^1 = x + y \end{align}

(7)

\begin{align} \quad (x + y)^2 = x^2 + 2xy + y^2 \end{align}

(8)

\begin{align} \quad (x+y)^3 = x^3 + 3x^2 y + 3xy^2 + y^3 \end{align}

The following theorem known as the Binomial Theorem states this result.

Theorem 1 (The Binomial Theorem): For all $x, y \in \mathbb{R}$ and $n \in \{ 0, 1, 2, ... \}$ we have that $\displaystyle{(x + y)^n = \binom{n}{0} x^ny^0 + \binom{n}{1} x^{n-1}y^1 + ... + \binom{n}{n-1} x^1y^{n-1} + \binom{n}{n} x^0y^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k}$.

• Proof: Let $n \in \{ 0, 1, 2, ... \}$ and considered the simplified expanded product of:

(9)

\begin{align} \quad (x + y)^n = \underbrace{(x + y)(x + y) ... (x + y)}_{n \:\mathrm{-factors}} \end{align}

• For each factor $(x + y)$ we obtain a term from distributing $x$ and another term from distributing $y$ across either the $x$ or the $y$ in the remaining factors, $(x + y)$. There are $\binom{n}{0} = 1$ ways to get the term $x^ny^0$ (by first multiplying all $x$'s in each of the $(x + y)$ factors), there are $\binom{n}{1} = n$ ways to get the term $x^{n-1}y^1$ (by multiplying any $n-1$ out of $n$ of the $x$'s and then the remaining $y$), etc… and so forth. By continuing in this fashion, we will eventually obtain all terms of the expansion of $(x + y)^n$ and get:

(10)

\begin{align} \quad (x + y)^n = \binom{n}{0} x^ny^0 + \binom{n}{1} x^{n-1}y^1 + ... + \binom{n}{n-1} x^1y^{n-1} + \binom{n}{n} x^0y^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k \quad \blacksquare \end{align}

The argument made in Theorem 1 is a bit subtle but nevertheless important. For a simpler example, consider the following expansion:

(11)

\begin{align} \quad (x + y)^3 = (x+y)(x+y)(x+y) \end{align}

Let $A= \{ x, y \}$ the set of terms in the factors $(x + y)$.

For each of $3$ multiplications, we will get a term corresponding to choosing one of the elements in $A$, and multiplying them together, that is, we get a finite sequence $\{ a, b, c \}$ where $a, b, c \in A$ and such that $abc$ is a term in the expansion of $(x + y)^3$. We thus choose to multiply either $0$, $1$, $2$, or $3$ of the $y$'s, and correspondingly, choose to multiply either $3$, $2$, $1$, or $0$ of the $x$'s. Thus the term $x^{n-k}y^k$ appears precisely $\binom{n}{k}$ times, and the full expansion of $(x + y)^3$ is:

(12)

\begin{align} \quad (x + y)^3 = x^3 + 3x^2y + 3xy^2 + y^3 = \sum_{k=0}^{3} \binom{3}{k} x^{n-k}y^k \end{align}

Before we end this page, recall that the binomial coefficients and Pascal's triangle are very much related. We have already noted and proven that a sequence of consecutive binomial coefficients in any row of Pascal's triangle is unimodal symmetric. We will now further show that the sum all binomial coefficients $\binom{n}{k}$ where $k$ is odd is equal to the sum of all binomial coefficients where $k$ is even for any row $n$.

Corollary 1: If $n$ and $k$ are positive integers that satisfy $0 \leq k \leq n$ then $\displaystyle{\sum_{k \: \mathrm{is \: odd}} \binom{n}{k} = \sum_{k \: \mathrm{is \: even}} \binom{n}{k}}$.

• Proof: Consider the binomial $(x + y)^n$. By Theorem 1 we have that the expansion of $(x + y)^n$ is given by:

(13)

\begin{align} \quad (x + y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k \end{align}

• Setting $x = 1$ and $y = -1$ gives us:

(14)

\begin{align} \quad (1 - 1)^n = \sum_{k=0}^{n} \binom{n}{k} 1^{n-k} (-1)^k \\ \quad 0 = \sum_{k=0}^{n} \binom{n}{k} (-1)^k \\ \quad 0 = \sum_{k \: \mathrm{is \: even}} \binom{n}{k} - \sum_{k \: \mathrm{is \: odd}} \binom{n}{k} \\ \quad \sum_{k \: \mathrm{is \: odd}} \binom{n}{k} = \sum_{k \: \mathrm{is \: even}} \binom{n}{k} \quad \blacksquare \end{align}