Binomial expansion: $(1+x)^n$

In a nutshell

The binomial expansion formula can be generalised to allow for the computation of expressions of the form $(1+x)^n$ when $n$ is any real number, and $|x| < 1$ via an infinite series.

Generalised binomial expansion formula

When $n$ is a natural number, the binomial expansion of $(a + b)^n$ is given by the formula:

$(a+b)^n = \binom{n}{0}a^n + \binom{n}{1}a^{n-1}b + \dots + \binom{n}{r}a^{n-r}b^r + \dots + \binom{n}{n}b^n$

Recall that $\binom{n}{r} = \dfrac{n!}{r!(n-r)!} = \dfrac{n \times (n-1) \times \dots \times (n-r+1)}{r!}$ . When $n$ is a natural number, and $r>n$ , the expression $\binom{n}{r} = 0$ . In the case that $n$ is not a natural number however, these binomial coefficients are never equal to $0$ for any natural number $r$ . Consequently, the expansion can be continued in order to obtain an infinite series:

$(1+x)^n = 1 + nx + \dfrac{n(n-1)}{2!}x^2 + \dfrac{n(n-1)(n-2)}{3!}x^3\dots + \frac{n(n-1)(n-2)\dots(n-r+1)}{r!}x^r + \dots$

This expansion is valid for any $n \in \mathbb{R}$ whenever $|x| < 1$ .

This infinite series can be used to find a valid expansion for $(1+bx)^n$ when $|bx| <1$ , i.e. when $|x|<\dfrac{1} {|b|}$ .

Example 1

Find the binomial expansion of $(1+x)^5$ where $|x| < 1$ .

By following the formula given above, you have

$\begin{aligned}(1+x)^5&=\binom{5}01^5x^0+\binom{5}{1}1^{4}x^1+\binom{5}{2}1^{3}x^2+\binom{5}{3}1^{2}x^3+\binom{5}{4}1^{1}x^4+\binom{5}{5}1^{0}x^5\\&=\underline{1+5x+10x^2+10x^3+5x^4+x^5}\end{aligned}$

Example 2

Find the expansion up to and including the $x^3$ term of the following expression:

$\dfrac{\left(^5\sqrt{1 + x}\right)^4}{\left(1 + x\right)}$

For what values of $x$ is the expansion valid?

The expression can be simplified to read:

$\begin{aligned}\dfrac{\left(1 + x\right)^{\frac45}}{\left(1 + x\right)} &= \left(1 + x\right)^{\frac45-1}\\&= \left(1 + x\right)^{-\frac15}\end{aligned}$ 

The generalised binomial expansion formula allows you to compute the expansion of this:

$1-\dfrac15x+\dfrac{\left(-\dfrac{1}{5}\right)\left(-\dfrac{6}{5}\right)}{2!}x^2 + \dfrac{\left(-\dfrac15\right)\left(-\dfrac65\right)\left(-\dfrac{11}{5}\right)}{3!}x^3 + \dots$ 

$= 1 - \dfrac{1}{5}x + \dfrac{3}{25}x^2 - \dfrac{11}{125}x^3 + \dots$ 

The expansion is valid when $\left|x\right| < 1$ .

Therefore, the expansion for $\dfrac{\left(^5\sqrt{1 + x}\right)^4}{\left(1 + x\right)}$ up to the $x^3$ term is $\underline{1 - \dfrac{1}{5}x + \dfrac{3}{25}x^2 - \dfrac{11}{125}x^3 + \dots}$ and is valid for $\underline{|x| < 1}$ .