# Binomial Expansion

Consider the following expansion of $$(x+a)^n$$

when  $$n=0;\quad (x+a)^{ 0 }=1$$

when  $$n=1;\quad (x+a)^{ 1 }=x+a$$

when  $$n=2;\quad (x+a)^{ 2 }=x^{ 2 }+2xa+a^{ 2 }$$

when  $$n=3;\quad (x+a)^{ 3 }=a^{ 3 }+3x^{ 2 }a+3xa^{ 2 }+a^{ 3 }$$

when  $$n=4;\quad (x+a)^{ 4 }=x^{ 4 }+4x^{ 3 }a+6x^{ 2 }a^{ 2 }+4xa^{ 3 }+a^{ 4 }$$

Observations from the above expansion are,

There are terms in the expansion of $$(x+a)^n$$

The power of the first term , start with n and reduce down to 0

The power of the second term a, start with 0 and increase up to n .

Coefficients of each term in the expansion can be obtained from the below Pascal’s triangle numbers.

Pascal Triangle

The general from of the expansion of $$(x+a)^n$$

$$(x+a)^{ n }=\left( \begin{matrix} n \\ 0 \end{matrix} \right) { x }^{ n }{ a }^{ 0 }+\left( \begin{matrix} n \\ 1 \end{matrix} \right) { x }^{ n-1 }{ a }^{ 1 }+\left( \begin{matrix} n \\ 2 \end{matrix} \right) { x }^{ n-2 }{ a }^{ 2 }+…..+\left( \begin{matrix} n \\ r \end{matrix} \right) { x }^{ n-r }{ a }^{ r }+…+\left( \begin{matrix} n \\ n-1 \end{matrix} \right) { x }^{ n-(n-1) }{ a }^{ n-1 }+\left( \begin{matrix} n \\ n \end{matrix} \right) { x }^{ n-n }{ a }^{ n }\quad \quad$$

Simplifying we get,
$$(x+a)^{ n }={ x }^{ n }{ a }+\left( \begin{matrix} n \\ 1 \end{matrix} \right) { x }^{ n-1 }{ a }^{ 1 }+\left( \begin{matrix} n \\ 2 \end{matrix} \right) { x }^{ n-2 }{ a }^{ 2 }+.\quad .\quad .\quad .\quad .+\left( \begin{matrix} n \\ r \end{matrix} \right) { x }^{ n-r }{ a }^{ r }+.\quad .\quad .+\left( \begin{matrix} n \\ n-1 \end{matrix} \right) { x }^{ 1 }{ a }^{ n-1 }+{ a }^{ n }\quad \quad$$

The general term in the expansion of$$(x+a)^n$$ is   $${ t }_{ r+1 }=\left( \begin{matrix} n \\ r \end{matrix} \right) { x }^{ n-r }{ a }^{ r }$$

Some results with $$\left( \begin{matrix} n \\ r \end{matrix} \right)$$

$$n!=n×(n-1)×(n-2)×……..×3×2×1$$

$$n! = n×(n-1)!$$

$$0! = 1 ; 1! = 1$$

$$\left( \begin{matrix} n \\ 0 \end{matrix} \right) =\left( \begin{matrix} n \\ n \end{matrix} \right) =1$$

$$\left( \begin{matrix} n \\ r \end{matrix} \right) =\frac { n! }{ (r!×(n-r)!) }$$

$$\left( \begin{matrix} n \\ r \end{matrix} \right) =\left( \begin{matrix} n \\ n-r \end{matrix} \right)$$

Some results with $$(1+x)^n$$

When a = 1 in the expansion of $$(x+a)^n$$we get $$(1+x)^n$$

$$(1+x)^n = {n \choose 0}x^0 + {n \choose 1}x^1 + {n \choose 2}x^2 + \cdots + {n \choose {n-1}}x^{n-1} + {n \choose n}x^n,$$

When $$n=\frac { 1 }{ 2 }$$, we get $$\sqrt{1+x}$$

$$\sqrt{1+x} = \textstyle 1 + \frac{1}{2}x – \frac{1}{8}x^2 + \frac{1}{16}x^3 – \frac{5}{128}x^4 + \frac{7}{256}x^5 – \cdots$$

When n = -1, we get $$(1+x)^{-1}$$

$$(1+x)^{-1} = \frac{1}{1+x} = 1 – x + x^2 – x^3 + x^4 – x^5 + \cdots$$

When $$n=\frac { -1 }{ 2 }$$ , we get $$(1+x)^\frac { -1 }{ 2 }$$

$$\frac{1}{\sqrt{1+x}} = \textstyle 1 -\frac{1}{2}x + \frac{3}{8}x^2 – \frac{5}{16}x^3 + \frac{35}{128}x^4 – \frac{63}{256}x^5 + \cdots$$

Practice Problems

(Total 5 marks)