Binomial Expansion

A binomial expansion is a method used to allow us to expand and simplify algebraic expressions in the form \( (x+y)^n\) into a sum of terms of the form \(ax^by^c\). If \(n\) is an integer, \(b\) and \(c\) also will be integers, and \(b + c = n\).

We can expand expressions in the form \( (x+y)^n\) by multiplying out every single bracket, but this might be very long and tedious for high values of \(n\) such as in \( (x+y)^{20}\) for example. This is where using the Binomial Theorem comes in useful.

The binomial theorem

The binomial theorem allows us to expand an expression of the form \( (x+y)^n\) into a sum. A general formula for a binomial expression is:

\[ (x+y)^n = \binom{n}{0}x^ny^0 + \binom{n}{1}x^{n-1}y^1 + \binom{n}{2}x^{n-2}y^2 + \dots + \binom{n}{n-1}x^1y^{n-1} + \binom{n}{n}x^0y^n.\]

Which can be simplified to:

\[ \begin{align} (x+y)^n &= \sum\limits_{k=0}^n \binom{n}{k} x^{n-k}y^k \\ &= \sum\limits_{k=0}^n \binom{n}{k} x^ky^{n-k} . \end{align}\]

Where both \(n\) and \(k\) are integers. This is also known as the binomial formula. The notation

\[ \binom{n}{k}\]

can be referred to as '\(n\) choose \(k\)' and gives a number called the binomial coefficient which is the number of different combinations of ordering \(k\) objects out of a total of \(n\) objects. The equation for the binomial coefficient (\(n\) choose \(k\) or \(^nC_r\) on a calculator) is given by:

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

Where '!' means factorial. Factorial means the product of an integer with all the integers below it. For example for \(5\) choose \(3\), we would have:

\[ \begin{align} \binom{5}{3} &= \frac{5!}{3!(5-3)!} \\ &= \frac{5\cdot 4\cdot 3 \cdot 2 \cdot 1}{(3\cdot 2\cdot 1)(2\cdot 1)} \\ &= 10. \end{align}\]

How do you do a binomial expansion?

To understand how to do a binomial expansion, we will look at an example. Let's say we want to expand \( (x+y)^4\). In this case, \(n = 4\) and \(k\) will vary between \(0\) and \(4\). Using the formula for the binomial expansion, we can write:

\[ (x+y)^4 = \binom{4}{0}x^4y^0 + \binom{4}{1}x^3y^1 + \binom{4}{2}x^2y^2 + \binom{4}{3}x^1y^3+\binom{4}{4}x^0y^4.\]

We can now use the equation for the binomial coefficient to find all the constant terms in this expression. For the first term we have:

\[ \begin{align} \binom{4}{0} &= \frac{4!}{0!(4-0)!} \\ &= \frac{4 \cdot 3\cdot 2\cdot 1}{1\cdot (4 \cdot 3\cdot 2\cdot 1 )} \\ &= 1. \end{align} \]

Repeating this for all five coefficients, we end up with binomial coefficients of \(1\), \(4\), \(6\), \(4\), \(1\) in order. Therefore, our expression for the binomial expansion simplifies to:

\[ x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + y^4.\]

Binomial expansion formula

To summarise the above explanation, the expansion formula can be written as:

\[(x+y)^n = \sum _{k=0}^{n} \binom{n}{k}x^{n-k}y^k = \sum _{k=0}^{n} \binom{n}{k}x^{k}y^{n-k}\]

Where \(\binom{n}{k}\) is the binomial coefficient of each term.

Binomial expansion for fractional and negative powers

Sometimes you will encounter algebraic expressions where n is not a positive integer but a negative integer or a fraction. Let's consider the expression \(\sqrt{1-2x}\) which can also be written as

\[ (1- 2x)^\dfrac{1}{2} \] where \(x < 0.5\). In this case, it becomes hard to find the formula to find the binomial coefficients,

because we can't find the factorials for a negative or rational number. However, if we look at an example for a positive integer, we can find a more general expression that we can then also apply to negative and fractional numbers. For example for

\[ \binom{6}{3}\]

we have

\[ \begin{align} \binom{6}{3}&= \frac{6!}{3!(6-3)!} \\ &= \frac{6\cdot 5\cdot 4}{3!} \\ &= \frac{6(6-1)(6-2)}{3!}. \end{align}\]

From this we observe that

\[ \binom{n}{k} = \frac{n(n-1)(n-2)(n-3)\dots (n-k+1)}{k!} \]

and therefore the more general expression for the binomial theorem is the infinite formula

\[ (a+b)^n = \frac{a^n}{0!} + \frac{na^{n-1}b}{1!} + \frac{n(n-1)a^{n-2}b^2}{2!} + \frac{n(n-1)(n-2)a^{n-3}b^3}{3!} + \dots \]

Let's look at \(\sqrt{1-2x}\). In this case \(a = -2x\), \(b = 1\) and \(n =1/2\). Substituting this we get:

\[ \begin{align} \frac{(-2x)^\frac{1}{2}}{0!} &+ \frac{\left(-\frac{1}{2}\right) (-2x)^{-\frac{1}{2}}\cdot 1 }{1!} \\ &\quad + \frac{\left(-\frac{1}{2}\right) \left(-\frac{1}{2}\right) (-2x)^{-\frac{3}{2}}\cdot 1^2 }{2!} \\ &\quad + \frac{\left(-\frac{1}{2}\right) \left(-\frac{1}{2}\right) \left(-\frac{3}{2}\right) (-2x)^{-\frac{5}{2}}\cdot 1^3 }{3!} + \dots \end{align}\]

Binomial expansion questions

We have collected a few questions with step-by-step solutions to help you understand how the binomial theorem and binomial expansion can be applied or asked about in an exam.