Binomial Expansion

What is the Binomial Expansion?

The binomial expansion gives a method for expanding a two-term expression in a bracket raised to a power
- For example ${(a + b)}^{n}$
- You may also see it referred to as the binomial theorem
- In this note $n$ will be a positive integer
  - See the 'General Binomial Expansion' revision note for the general case
To expand a bracket with a two-term expression in it:
- Determine what $a$ and $b$ are for your example
- Then use the formula for the binomial expansion
  - ${(a + b)}^{n} = a^{n} + (\begin{matrix} n \\ 1 \end{matrix}) a^{n - 1} b + \dots + (\begin{matrix} n \\ r \end{matrix}) a^{n - r} b^{r} + \dots + b^{n}$

- in the formula is known as the binomial coefficient
  - $(\begin{matrix} n \\ r \end{matrix}) = \frac{n!}{r! (n - r)!} = \frac{n (n - 1) . . . (n - r + 1)}{r!}$
  - $n!$ (" $n$ factorial") is defined by $n! = 1 \times 2 \times 3 \times . . . \times n$
  - You may also see $(\begin{matrix} n \\ r \end{matrix})$ written as ${}^{n}{C_{r}}$
  - Your calculator should be able to calculate $(\begin{matrix} n \\ r \end{matrix})$ for you
  - Or you can use Pascal's triangle (see the next section)
- To get all the terms
  - Start with $r = 0$
  - Then use $r = 1$ , $r = 2$ ,... until you get up to $r = n$
  - So there will always be $n + 1$ terms in the full expansion
- This version of the binomial expansion formula is not on the exam formula sheet
  - But it is a special case of the Binomial Series formula for ${(1 + x)}^{n}$ which is on the formula sheet
  - See the 'General Binomial Expansion' revision note
When expanding something like ${(p + q x)}^{n}$ you may only be asked to find the first few terms of an expansion
- Check whether the question wants ascending or descending powers of x
  - For ascending powers start with the constant term, $p^{n}$
  - For descending powers start with the term with $x$ , ${(q x)}^{n}$
  - Choosing $a$ and $b$ appropriately will make it easier to follow the formula above
If you are not writing the full expansion you can either
- show that the series continues by putting an ellipsis (…) after your final term
- or show that the terms you have found are an approximation of the full series by using the 'approximately equals' sign (≈)

Finding binomial coefficients using Pascal's triangle

Pascal’s triangle is a way of arranging (and finding!) the binomial coefficients
- The first row has just the number 1
- Each row begins and ends with a 1
- Starting in the third row
  - Each other terms is the sum of the two terms immediately above it

Pascal's Triangle

Pascal’s triangle is an alternative way of finding the binomial coefficients $(\begin{matrix} n \\ r \end{matrix})$ (also written ${}^{n}{C_{r}}$ )
- It can be useful for finding the values of the coefficients without a calculator
  - Most useful for smaller values of $n$
  - For larger values of $n$ it is slow and prone to arithmetic errors

Taking the first row as corresponding to $n = 0$ ,
- each row gives the binomial coefficient values for the corresponding value of $n$
- within a row the values run from $r = 0$ to $r = n$
- e.g. from the 6^th row of the table ():
  - ${(a + b)}^{5} = a^{5} + 5 a^{4} b + 10 a^{3} b^{2} + 10 a^{2} b^{3} + 5 a b^{4} + b^{5}$

How do I find the coefficient of a single term?

You may just be asked to find the coefficient of a single term, rather than the whole expansion
Use the formula for the general term
- $(\begin{matrix} n \\ r \end{matrix}) a^{n - r} b^{r}$

To find a particular power of $x$ term in an expansion
- Choose which value of $r$ you will need to use in the formula
- The laws of indices can help you decide which value of $r$ to use:
  - For ${(p + q x)}^{n}$ , to find the coefficient of $x^{2}$ let $a = p, b = q x$ and use $r = 2$
  - For ${(p + q x^{2})}^{n}$ , to find the coefficient of $x^{2}$ let $a = p, b = q x^{2}$ and use $r = 1$
  - For something like ${(p x + \frac{q}{x})}^{n}$ , you need to consider how the powers will cancel each other
    - E.g. for ${(p x + \frac{q}{x})}^{6}$ , to find the coefficient of $x^{2}$ let $a = p x, b = \frac{q}{x}$ and use $r = 2$
    - Because then $x^{n - r} {(\frac{1}{x})}^{r} = x^{6 - 2} {(\frac{1}{x})}^{2} = x^{4} (\frac{1}{x^{2}}) = x^{2}$
  - There are a lot of variations, so practice is better than trying to memorise formulae for $r$ !
If you know the coefficient of a particular term, you can use it to find an unknown in the brackets
- Use the laws of indices to choose the correct term
- Then use the general term formula to form and solve an equation

Exam Tip

Binomial expansion questions can get messy
- Use separate lines to keep your working clear
- And always put terms in brackets

Worked example

Using the binomial expansion, find the complete expansion of ${(x + y)}^{4}$ .

Use the formula with $a = x$ , $b = y$ and $n = 4$
$r$ will run from 0 to 4, so there will be 5 terms

${(x + y)}^{4} = (\begin{matrix} 4 \\ 0 \end{matrix}) x^{4} + (\begin{matrix} 4 \\ 1 \end{matrix}) x^{3} y + (\begin{matrix} 4 \\ 2 \end{matrix}) x^{2} y^{2} + (\begin{matrix} 4 \\ 3 \end{matrix}) x y^{3} + (\begin{matrix} 4 \\ 4 \end{matrix}) y^{4}$

Now just work out the values of the binomial coefficients
You can use the formula, your calculator or Pascal's triangle

Note that $(\begin{matrix} 4 \\ 0 \end{matrix}) = (\begin{matrix} 4 \\ 4 \end{matrix}) = 1$
That's why we usually don't bother writing the binomial coefficients for the first and last terms of an expansion!

${(x + y)}^{4} = x^{4} + 4 x^{3} y + 6 x^{2} y^{2} + 4 x y^{3} + y^{4}$

Worked example

Find the first three terms, in ascending powers of $x$ , in the expansion of $(3 - 2 x)^{5}$ .

For ascending powers of $x$ we want to start with the constant term

So we want to use the formula with $a = 3$ , $b = - 2 x$ , and $n = 5$

For the first three terms (constant term, $x$ term and $x^{2}$ term) we want $r$ from 0 to 2

Substitute those values into the formula

${(3 - 2 x)}^{5} = {(3)}^{5} + (\begin{matrix} 5 \\ 1 \end{matrix}) {(3)}^{4} (- 2 x) + (\begin{matrix} 5 \\ 2 \end{matrix}) {(3)}^{3} {(- 2 x)}^{2} + . . .$

Find the value of the binomial coefficients and bring the powers inside the brackets
Be careful with the minus signs!

${(3 - 2 x)}^{5} = 243 + (5) (81) (- 2 x) + (10) (27) (4 x^{2}) + . . .$

Expand the remaining brackets and write down the final answer

${(3 - 2 x)}^{5} = 243 - 810 x + 1080 x^{2} + . . .$

Binomial Expansion (Edexcel IGCSE Further Maths)

Revision Note