# Binomial Theorem

## Factorials

A factorial function is a function that multiplies first $n$ natural numbers. For a natural number $n$ with $n!$ we denote the multiplication of first $n$ natural numbers.

For example:

$$1! = 1,$$
$$2! = 1 \cdot 2 = 2,$$
$$3! = 1 \cdot 2 \cdot 3 = 6,$$
$$4! = 1 \cdot2 \cdot 3 \cdot4 = 24,$$
$$5! = 1\cdot 2 \cdot 3 \cdot4 \cdot5 = 120.$$

In addition, it is useful to define the value $0!.$ We define $0!:=1$ by agreement.

By inductive conclusion, we can see that the factorials match the formula

$$n! = n \cdot (n – 1)!$$

with initial value $0!=1.$

Example 1. Calculate the following expression:

$$\frac{49! – 48!}{48!}.$$

Solution:

As a tip, when dealing with factorials never jump ahead and calculate everything because you can always somehow use the recursion. Now, since we know that $49! = 49 \cdot48!$ we have a common factor of all terms:

$$\frac{49 \cdot48! – 48!}{48!} = \frac{48! (49 – 1)}{48!} = 48.$$

Example 2. Solve the following equation:

$$\frac{n!}{(n – 3)!} = \frac{8 (n – 1)!}{(n – 2)!}.$$

Solution:

$$\frac{n!}{(n – 3)!} = \frac{8 (n – 1)!}{(n – 2)!}$$

$$\Leftrightarrow \frac{n \cdot (n – 1) \cdot(n – 2) \cdot(n – 3)!}{(n – 3)!} = \frac{8 (n – 1) (n – 2)!}{(n – 2)!}$$

$$\Leftrightarrow n \cdot (n – 1) \cdot (n – 2) = \ 8 \cdot(n – 1)$$

$$\Leftrightarrow n \cdot(n – 2) = 8$$

$$\Leftrightarrow n^2 – 2n – 8 = 0$$

$$\Rightarrow n_{1,2} = \frac{2\pm \sqrt{2^{2}+ 4\cdot 1\cdot 8}}{2\cdot 1}$$

$$\Rightarrow n_{1,2} =\frac{2\pm \sqrt{4+32}}{2}$$

$$\Rightarrow n_{1,2} =\frac{2\pm \sqrt{36}}{2}$$

$$\Rightarrow n_{1,2} =\frac{2\pm 6}{2}$$

$$\Rightarrow n= 4 \quad \textrm{or} \quad n =- 2$$

$n=4$ is the only solution to the default equation because $n$ must be a natural number.

## Binomial coefficients

Let both $n$ and $k$ are natural numbers and if $k$ can also have the value of $0$ and $k \le n.$ The binomial coefficient is denoted with symbol ${{n}\choose{k}}$ and defined as:

$${{n}\choose{k}} :=\frac{n!}{k! (n – k)!},$$ for $k\geq 1.$ For $k=0$ by definition, we have:

$${{n}\choose{0}}:=1.$$

When replacing $k$ with $0$ in the definition of the binomial coefficient we get:

$${{n}\choose{0}} = \frac{n!}{0! (n – 0!)} = \frac{n!}{1 \cdot n!} = \frac{n!}{n!} = 1.$$

## The symmetry property

$${{n}\choose{k}} = {{n}\choose{n – k}}, \qquad k = 0, 1, 2, …, n.$$

Proof.

$${{n}\choose{n-k}} = \frac{n!}{(n-k)![n-(n-k)]!} =\frac{n!}{(n-k)!k!}={{n}\choose{k}}.$$

Example 3. Calculate ${8}\choose{6}$ using the symmetry property as above.

Solution:

$${{8}\choose{6}} = {{8}\choose{8 – 6}} = {{8}\choose{2}} = \frac{8 \cdot 7}{2 \cdot 1} = 28.$$

## Pascal’s triangle

Let’s count the binomial coefficients for small value numbers $n$ and their calculated values as shown below:

$${{1}\choose{0}}\qquad {{1}\choose{1}}$$

$${{2}\choose{0}}\qquad {{2}\choose{1}}\qquad {{2}\choose{2}}$$

$${{3}\choose{0}} \qquad {{3}\choose{1}}\qquad {{3}\choose{2}}\qquad {{3}\choose{3}}$$

$${{4}\choose{0}}\qquad {{4}\choose{1}}\qquad {{4}\choose{2}}\qquad {{4}\choose{3}} \qquad{{4}\choose{4}}$$

$$\vdots$$

The triangle above is commonly known as a Pascal’s or Chinese triangle. Write the calculated values with two more added lines:

$$1 \qquad 1$$

$$1 \qquad 2 \qquad 1$$

$$1 \qquad 3 \qquad 3 \qquad 1$$

$$1 \qquad 4 \qquad 6 \qquad 4 \qquad 1$$

$$1 \qquad 5 \qquad 10 \qquad 10 \qquad 5 \qquad 1$$

$$1 \qquad 6 \qquad 15 \qquad 20 \qquad 15 \qquad 6 \qquad 1$$

$$\vdots$$

Each element from the Pascal’s triangle is equal to the addition of two elements on the line above on the right and left side of applicable element, except the constant value of the elements on the edges of the triangle which are always equal to $1.$

This property is valid for any element of Pascal’s triangle. Choose three characteristic elements of Pascal’s triangle:

$${{n}\choose{k-1}} \qquad {{n}\choose{k}}$$

$${{n+1}\choose{k}}$$

The following relation describes the basic principle of Pascal’s triangle:

$${{n}\choose{k-1}}+{{n}\choose{k}}={{n+1}\choose{k}}.$$

Proof.

$${{n}\choose{k-1}}+{{n}\choose{k}}=$$

$$\frac{n!}{(k-1)![n-(k-1)]!} + \frac {n!}{k!(n-k)!}=\frac{k\cdot n! + (n-k+1)\cdot n!}{k!(n-k+1)!} =\frac{n!(k+n-k+1)}{k!(n+1-k)!}$$

$$= \frac{n!(n+1)}{k!(n+1-k)!}=\frac {(n+1)!}{k!(n+1-k)!}$$

$$={{n+1}\choose{k}}.$$

It has been proven that the left side of the equal sign is equal to the right.

This property demonstrates that all binomial coefficients are natural numbers because each of them is equal to $1$ or to the sum of two natural numbers.

## The binomial theorem

The binomial theorem, is also known as binomial expansion, which explains the expansion of powers. It only applies to binomials. Let’s take a look at the link between values in Pascal’s triangle and the display of the powers of the binomial $(a+b)^n.$

For small values of a natural number $n$ we know the formulas for the powers of the binomial:

$$(a + b)^1 = 1 \cdot a + 1 \cdot b,$$

$$(a + b)^2 = 1 \cdot a^2 + 2 \cdot a \cdot b + 1\cdot b^2,$$

$$(a + b)^3 = 1 \cdot a^3 + 3 \cdot a^2 \cdot b + 3 \cdot a \cdot b^2 + 1 \cdot b^3.$$

The last two formulas are called square of a binomial and cube of a binomial, respectively. The coefficients in these formulas are in fact binomial coefficients.

If we continue further, we get:

$$(a+b)^4 = a^4+4\cdot a^3 \cdot b + 6\cdot a^2\cdot b^2 + 4\cdot a \cdot b^3 + b^4,$$

$$(a+b)^5 = a^5 +5\cdot a^4\cdot b + 10\cdot a^3 \cdot b^2 + 10\cdot a^2 \cdot b^3 + 5\cdot a \cdot b^4 + b^5$$

$$\vdots$$

Notice that all addends come in the form $b_i \cdot a^{n – i} \cdot b^i$. Numbers $b_i$ are called binomial coefficients. They are easily calculated and noted using factorials.

The binomial theorem

For $\forall$ $a, b \in \mathbb{R},$ $n\in \mathbb{N}$ is valid:

$$(a+b)^n = {{n}\choose{0}} a^{n} b^{0} + {{n}\choose{1}} a^{n-1} b^{1} + {{n}\choose{2}} a^{n-2} b^{2}+ \cdots + {{n}\choose{n-1}} a^{1} b^{n-1} + {{n}\choose{n}} a^{0} b^{n}.$$

Example 4. Through using the above formula expand the following:

$$(2x + 1) ^6.$$

Solution:

$$(2x+1)^6 = {{6}\choose{0}} (2x)^{6}1^{0} + {{6}\choose{1}} (2x)^{5} 1^{1} + {{6}\choose{2}} (2x)^{4} 1^{2}+ {{6}\choose{3}} (2x)^{3} 1^{3} + {{6}\choose{4}} (2x)^{2} 1^{4} +{{6}\choose{5}} (2x)^{1} 1^{5} + {{6}\choose{6}} (2x)^{0} 1^{6}$$

$$=1\cdot 64 x ^{6}\cdot 1 + 6\cdot 32 x^{5} \cdot 1 + 15 \cdot 16 x^{4}\cdot 1 + 20 \cdot 8 x^{3}\cdot 1 + 15 \cdot 4x^{2}\cdot 1 + 6 \cdot 2 x^\cdot 1 + 1\cdot 1\cdot 1$$

$$=64 x ^{6} + 192x^{5} + 240x^{4} + 160x^{3} + 60x^{2} + 12x + 1.$$