Problem Solving with Binomial Expansion

How do I find a specific term of a binomial expansion?

You may be asked to find a specific term or coefficient of a term in an expansion, rather than finding the entire expansion
We can use several facts to help us
- The powers in each term sum to $n$
- The coefficients for ${(a + b)}^{n}$ come from the $n^{t h}$ row of Pascal's triangle
- The coefficient of $a^{m}$ will be the $m^{t h}$ number in that row (and also the ${(n - m)}^{t h}$ number in that row)
  - e.g. In ${(a + b)}^{7}$ the coefficient of $a^{2}$ will be the 2^nd number in the 7^th row (or the 5^th number in the 7^th row)
- Note that this counting system includes the first number in each row
  - So the 1^st number (of every row) is 1
  - This was previously called the 0^th number in the row
  - However patterns are easier to see this way for the purposes of problem solving
For example to find the coefficient of $x^{4}$ in ${(2 x + 3)}^{6}$
- The powers in the term will sum to 6, and we need to include $x^{4}$ , so they must be
  - $C \times {(2 x)}^{4} \times {(3)}^{2}$
  - where $C$ is the coefficient from Pascal's triangle
- $n = 6$ and we are looking at the term with a power of 4 (and 2)
  - So we will use the 4^th (or 2^nd) number in the 6^th row of Pascal's triangle, which is 15
- So the term is $15 \times {(2 x)}^{4} \times {(3)}^{2}$ which we can then expand and simplify
- $15 \times 2^{4} x^{4} \times 9 = 2160 x^{4}$
- So the coefficient of $x^{4}$ is 2160

How do I find an unknown with binomial expansion?

Sometimes you may be asked to find an unknown in an expansion
For example, you may be told that for the expansion ${(p + 2 x)}^{5}$ , the coefficient of $x^{3}$ is 2000, and you can then use this information to find $p$
Use the method described above to find the specific term containing an $x^{3}$ , which will have its coefficient in terms of $p$
- Then equate this coefficient to 2000 and solve to find $p$

When do powers cancel out in a binomial expansion?

Sometimes the two terms in the binomial can cause powers to cancel out
For example when expanding ${(x + \frac{1}{x})}^{4}$ we would find
- ${(x + \frac{1}{x})}^{4} = x^{4} + 4 x^{3} (\frac{1}{x}) + 6 x^{2} {(\frac{1}{x})}^{2} + 4 x {(\frac{1}{x})}^{3} + {(\frac{1}{x})}^{4}$
- Which simplifies to
- $x^{4} + 4 x^{3} (\frac{1}{x}) + 6 x^{2} (\frac{1}{x^{2}}) + 4 x (\frac{1}{x^{3}}) + (\frac{1}{x^{4}})$
You can see that the powers in each term will change, as we have powers of $x$ divided by each other
- This expression will reduce to
- $x^{4} + 4 x^{2} + 6 + \frac{4}{x^{2}} + \frac{1}{x^{4}}$ or $x^{4} + 4 x^{2} + 6 + 4 x^{- 2} + x^{- 4}$
You can see that this expansion now looks quite different to "standard" expansions
- We now have a constant term in the middle, rather than at the end
For this reason you need to be careful with expansions that look like this
- They typically contain a multiple of $\frac{1}{x}$ or $\frac{1}{x^{2}}$ etc in the binomial
  - e.g. ${(2 x - \frac{3}{x})}^{n}$ or ${(4 x + \frac{2}{x^{2}})}^{n}$

Exam Tip

Look out for extra information about unknowns
- they will often say "...where $p > 0$ "
- this will help you if you get a solution like $p = \pm 3$ and you can then select the positive value
If you feel unsure about finding a specific term straight away, writing out the first few terms can help you remember or spot the pattern to use

Worked example

(a)

Find the coefficient of

x^{3}

in the expansion of

{(4 x - 2)}^{5}

$a = 4 x$ and $b = - 2$
For the term involving $x^{3}$ we will need the term involving ${(4 x)}^{3}$ and the 3^rd number in the 5^th row from Pascal's triangle

5^th row of Pascal's triangle is

$\begin{matrix} 1 & 5 & 10 & 10 & 5 & 1 \end{matrix}$

Therefore the term required is

$10 \times {(4 x)}^{3} \times {(- 2)}^{2}$

So the coefficient of $x^{3}$ is

$\begin{array}{rcl} 10 \times 4^{3} \times {(- 2)}^{2} & = & 10 \times 64 \times 4 \\ = & 2560 \end{array}$

The coefficient of the $x^{3}$ term is 2560

(b)

Given that $p > 0$ and that the coefficient of $x^{4}$ in the expansion of ${(3 x - p)}^{6}$ is 79 380, find the value of $p$ .

$a = 3 x$ and $b = - p$
For the term involving $x^{4}$ we will need the term involving ${(3 x)}^{4}$ and the 4^th number in the 6^th row from Pascal's triangle

The 6^th row of Pascal's triangle is

$\begin{matrix} 1 & 6 & 15 & 20 & 15 & 6 & 1 \end{matrix} \begin{matrix} \end{matrix}$

Therefore the term required is

$20 \times {(3 x)}^{4} \times {(- p)}^{2}$

So the coefficient of $x^{3}$ is

$\begin{array}{rcl} 20 \times 3^{4} \times {(- p)}^{2} & = & 20 \times 81 \times p^{2} \\ = & 1620 p^{2} \end{array}$

The question gives the coefficient as 79 380 so set up and solve an equation for $p$

$\begin{array}{rcl} 1620 p^{2} & = & 79 380 \\ p^{2} & = & 49 \end{array}$

$p > 0$ so the positive square root is needed

$p = 7$

(c)

\begin{matrix}  \end{matrix}

Find the coefficient of $x$ in the expansion of ${(2 x + \frac{1}{x})}^{5}$ .

Spot that this is a question where powers of $x$ will cancel
$a = 2 x$ and $b = \frac{1}{x}$
The $x$ term will be generated when the power of $(2 x)$ is one greater than the power of $(\frac{1}{x})$
As powers in each term will sum to 5 in this case, this will be the term involving ${(2 x)}^{3}$ (and ${(\frac{1}{x})}^{2}$

The 3rd number in the 5th row of Pascal's triangle is also needed

$\begin{matrix} 1 & 5 & 10 & 10 & 5 & 1 \end{matrix}$

Therefore the term required is

$10 \times {(2 x)}^{3} \times {(\frac{1}{x})}^{2}$

So the coefficient of $x$ is

$\begin{array}{rcl} 10 \times 2^{3} \times 1^{2} & = & 10 \times 8 \times 1 \\ = & 80 \end{array}$

The coefficient of the $x$ term is 80

Problem Solving with Binomial Expansion (AQA GCSE Further Maths)

Revision Note