Hello. This is a math post about the binomial formula. The binomial formula algebra technique allows us to expand binomials such as \((a + b)^{n}\) where \(n \geq 2\).

 

Table of Contents

 

 

Factorials

 

Before we get right into the binomial formula, we need the ingredients first. We start with the factorial.

Suppose we have something like \(3 \cdot 2 \cdot 1 = 6\) and \(5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 120\). We would like a more compact form of writing this (especially for larger numbers).

The factorial form is expressed as \(n!\) where \(n \geq 1\) and:

 

\[\displaystyle n! = n \cdot (n - 1) \cdot (n - 2) \text{ ... } \cdot 1\]

 

The alternative formula is:

 

\[\displaystyle n! = \prod_{k = 1}^{n} k = 1 \cdot 2 \cdot \text{ ... } \cdot (n-1) \cdot n$ We can express $3 \cdot 2 \cdot 1 = 6$ as $3! = 6$ and $5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 120$ as $5! = 120\].

 

Some Factorial Properties

\[n! = n \cdot (n -1)!\]

\[0! = 1\]

 

Binomial Coefficient

The next piece we need is the binomial coefficient. In terms of practical applications, this binomial coefficient represents the number of ways (combinations) of choosing \(k\) unordered items from \(n\) choices.

Mathematically, the binomial coefficient is represented as:

 

\[\binom{n}{k} = \dfrac{n!}{k! (n-k)!}\]

 

Some Properties

 

\(\binom{n}{n} = \binom{n}{0} = 1\) for integers \(n \geq 0\).

\(\binom{0}{0} = 1\) (Corollary/Special Case from the above)

 

Pascal’s Triangle

 

We now look at Pascal’s Triangle. Pascal’s triangle is related to the coefficients from the binomial formula. Refer to the image below.

Image Source

 

The numbers in Pascal’s triangle can also be expressed using binomial coefficients and the \(\binom{n}{k}\) notation.

 

The Binomial Formula

 

Now we have the necessary ingredients. Here is the binomial formula for \(n \geq 1\)

 

\[\begin{array} {lcl} \displaystyle (a + b)^{n} & = & \sum_{k = 0}^{n} \binom{n}{k} a^{n-k}b^{k} \\ & = & \binom{n}{0}a^{n} + \binom{n}{1} a^{n-1}b + \text{ ...} + \binom{n}{n} b^{n} \\ & = & a^{n} + na^{n-1}b + \text{ ...} + b^{n} \end{array}\]

 

Examples

 

Here are a few examples which apply the binomial formula.

  1. Example One: \((x + 2)^2\)

 

\[\displaystyle \begin{array} {lcl} (x + 2)^{2} & = & \sum_{k = 0}^{n = 2} \binom{2}{k} x^{2-k}2^{k}\\ & = & \binom{2}{0} x^{2}2^{0} + \binom{2}{1} x^{1}2^{1} + \binom{2}{2} x^{0}2^{2}\\ & = & 1\cdot x^2 + 2 \cdot 2x + 1 \cdot 4\\ & = & x^2 + 4x + 4\\ \end{array}\]

 

  1. Example Two: \((x - 5)^3\)

 

\[\displaystyle \begin{array} {lcl} (x - 5)^{3} & = & \sum_{k = 0}^{n = 3} \binom{3}{k} x^{3-k}(-5)^{k}\\ & = & \binom{3}{0} x^{3}(-5)^{0} + \binom{3}{1} x^{2}(-5)^{1} + \binom{3}{2} x^{1}(-5)^{2} + \binom{3}{3} x^{0}(-5)^{3}\\ & = & 1\cdot x^3 - 3 \cdot 5x^2 + 3 \cdot 25x - 1 \cdot 125\\ & = & x^3 - 15x^2 + 75x - 125\\ \end{array}\]

 

  1. Example Three: \((a - b)^4\)

 

\[\displaystyle \begin{array} {lcl} (a - b)^{4} & = & \sum_{k = 0}^{n = 4} \binom{4}{k} a^{4-k} (-b)^{k}\\ & = & \binom{4}{0} a^{4}(-b)^{0} + \binom{4}{1} a^{3}(-b)^{1} + \binom{4}{2} a^{2}(-b)^{2} + \binom{4}{3} a^{1}(-b)^{3} + \binom{4}{4} a^{0}(-b)^{4}\\ & = & 1 \cdot a^4 - 4 \cdot a^{3}b + 6 \cdot a^2 b^2 - 4ab^3 + 1 \cdot (-b)^{4}\\ & = & a^4 - 4 a^{3}b + 6 a^2 b^2 - 4ab^3 + b^{4}\\ \end{array}\]

 

Notice that with binomials with one positive term and one negative term the signs are alternating in the sum. This is seen in examples two and three.

The featured image is from http://exchangedownloads.smarttech.com/public/content/2b/2b616ad0-d9ab-496d-b82b-a2c2dc5dc381/previews/medium/0001.png.