# Binomial Expansion Formula – AS Level Maths

Are you needing help to power through advanced maths content on your own terms? Then you’ve come to the right place. This tuitional-style post on the binomial expansion formula for AS Level is a great spot to get immersed in the topic whilst untangling any misconceptions.

So, let’s get started with some definitions and clarification before seeing how this links to Pascal’s Triangle and, finally, walking through a series of worked examples.

### What is Binomial Expansion?

A binomial is an algebraic expression which contains two terms. A binomial expansion is when a binomial is raised to a power, for example:

$(1+x)^2$

$(2x+3y^2)^5$

$(5-2x)^{-3}$

Here, we will look at the binomial expansions in the form:

$(a+b)^{n}$

where n is a positive integer.

Click here for a selection of teaching resources on binomial expansions. This includes the information in this blog in printable or presentable forms, printable activities, an interactive activity, a prior knowledge test to prepare for the topic and binomial expansion exam-style questions to practise afterwards.

### Pascal’s Triangle

Consider the expressions $(a+b)^{0}$, $(a+b)^{1}$, $(a+b)^{2}$ and $(a+b)^{3}$ – each can be expanded as follows:

$(a+b)^{0} = \qquad \qquad \qquad \; 1$

$(a+b)^{1} = \qquad \quad \qquad 1a + 1b$

$(a+b)^{2} = \qquad \quad 1a^2 + 2ab + 1b^2$

$(a+b)^{3} = \qquad 1a^3 + 3a^{2}b + 3ab^{2} + 1b^3$

You might spot a pattern; the first and last terms have a coefficient of 1 and the remaining coefficients are found by adding the two coefficients immediately above that term.

In fact, the pattern in these coefficients can be extended. This is known as Pascal’s triangle:

$1$

$1 \qquad 1$

$1 \qquad 2 \qquad 1$

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

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

The $n^{\text{th}}$ row in Pascal’s triangle gives the coefficients of each term in the expansion of $(a+b)^{n-1}$.

Have another look at each expansion – another pattern is made by the powers in each term. The sum of the powers in each term is equal to the original power, where the power of a decreases by 1 each time and the power of b increases by 1 each time. For example, look at each term in the expansion of $(a+b)^{3}$:

$a^3: 3 + 0 = 3$

$a^2b: 2 + 1 = 3$

$ab^2: 1 + 2 = 3$

$b^3: 2 + 1 = 3$

By following this pattern, we can find the expansion of binomials with small, integer values of n.

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## Binomial Expansion Formula – AS Level Examples

Example Question 1:
Use Pascal’s triangle to find the expansion of $\boldsymbol{(2 + 4x)^3}$

Since the power is 3, we use the 4th row of Pascal’s triangle to find the coefficients: 1, 3, 3 and 1. We also know that the power of 2 will begin at 3 and decrease by 1 each time. Similarly, the power of 4x will begin at 0 and increase by 1 each time.

It can be sensible to structure this in a table and then multiply each column:

$\boldsymbol{(2 + 4x)^3 = 8 + 48x + 96x^2 + 64x^3}$

For large powers of n, it might be time consuming to write out Pascal’s triangle. In these cases, we can use the binomial expansion formula.

The expansion of $(a + b)^n$, for integer values of n, is given by:
$(a+b)^n = a^n + \binom{n}{1} a^{n-1}b + \binom{n}{2} a^{n-2} b^2 \\ \hspace*{15ex} + \binom{n}{3} a^{n-3}b^3 + ... + b^n$

Where:
$\binom{n}{r} = {}^n\text{C}_r = \frac{n!}{r!(n-r)!}$

And:
$n! = n \times (n-1) \times (n-2) \times ... \times 2 \times 1$

Example Question 2:
Find the binomial expansion of $\boldsymbol{(2-x)^5}$

$\text{Let } a = 2, b = -x \text{ and } n = 5$

$(2-x)^5 = 2^5 + \binom{5}{1} \times 2^4 \times (-x) + \binom{5}{2} \times 2^3 \times (-x)^2 \\ \hspace*{5ex} + \binom{5}{3} \times 2^2 \times (-x)^3 + \binom{5}{4} \times 2 \times (-x)^4 + (-x)^5$

$(2-x)^5 = 32 + 5 \times 16 \times (-x) + 10 \times 8 \times x^2 \\ \hspace*{5ex} + 10 \times 4 \times (-x)^3 + 5 \times 2 \times x^4 + (-x)^5$

$\boldsymbol{(2-x)^5 = 32 - 80x +80x^2 - 40x^3 +10x^4 - x^5}$

Example Question 3:
Find the constant term in the expansion of $\boldsymbol{(x + \frac{1}{x})^4}$

$\text{Let } a = x, b = \frac{1}{x} \text{ and } n = 4$

$(x + \frac{1}{x})^4 = x^4 + \binom{4}{1} \times x^3 \times \frac{1}{x} +\binom{4}{2} \times x^2 \times (\frac{1}{x})^2 \\ \hspace*{5ex}+ \binom{4}{3} \times x \times (\frac{1}{x})^3 + (\frac{1}{x})^4$

We can either complete the expansion, or we can notice that the variables only cancel in the middle term, leaving us with the constant term:

\begin{aligned} {}^4\text{C}_2 \times x^2 \times (\frac{1}{x})^2 &= 6 \times x^2 \times \frac{1}{x^2}\\ &= 6 \end{aligned}

The constant term is 6.

Another interesting use of the binomial expansion is to calculate estimations of powers. For values of x which are close to zero, we can ignore larger powers of n to approximate the power of a number.

Example Question 4:
Use the first three terms, in ascending powers of x, in the expansion of $\boldsymbol{(2 + x)^{10}}$ to estimate the value of 2.0310.

The first three terms are given by:

\begin{aligned} (2+x)^{10} &\approx 2^{10} + \binom{10}{1} \times 2^9 \times x + \binom{10}{2} \times 2^8 \times x^2 \\ &= 1024 + 5120x + 11520x^2 \end{aligned}

To find the value of x:

\begin{aligned} 2 + x &= 2.03 \\ x &= 0.03 \end{aligned}

We can then substitute x into the first three terms of the expansion:

\begin{aligned} 2.03^{10} &\approx 1024 + 5120 \times 0.03 + 11520 \times 0.03^2 \\ &= \boldsymbol{1187.968} \end{aligned}

The actual value of 2.0310 is 1188.393… so the approximation is correct to the nearest whole number.

Let’s also look at the final two terms in the full binomial expansion to show that they would have had little impact on the final answer:

\begin{aligned} \binom{10}{9} \times 2 \times x^9 \times x^{10} &= 10 \times 2 \times 0.3^9 \times 0.3^{10} \\ &=0.0003995649 \end{aligned}

This would have very little impact on the accuracy of the final estimate.

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