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 $$\underline{|x| < 1}$$.