General Binomial Expansion (Edexcel IGCSE Further Pure Maths)

Revision Note

Written by: Roger B

Reviewed by: Dan Finlay

General Binomial Expansion

What is the general binomial expansion?

The general binomial expansion lets us write ${(1 + x)}^{n}$ as a binomial series
- It is valid for any $n \in ℚ$
  - $ℚ$ is the set of all rational numbers
  - So $n$ can be negative or a fraction
- If $n$ is a positive integer
  - Then the series has a finite number of terms
  - For this case see the 'Binomial Expansion' revision note

If $n \in ℚ$ is not a positive integer
- Then the series has an infinite number of terms
- I.e. 'it goes on forever'
- An exam question will only ask for the first few terms of the expansion

A general binomial expansion is found using the binomial series formula
- ${(1 + x)}^{n} = 1 + n x + \frac{n (n - 1)}{2!} x^{2} + . . . + \frac{n (n - 1) . . . (n - r + 1)}{r!} x^{r} + . . . for |x| < 1, n \in ℚ$
- This formula is on the exam formula sheet
  - So you don't need to remember it
  - But you do need to know how to use it
The expansion is only valid for $|x| < 1$
- This means $- 1 < x < 1$
- This is known as the interval of convergence
- For values of $x$ inside the interval of convergence
  - the (infinite) expansion on the right-hand side of the formula
  - is exactly equal to the function on the left-hand side

How do I use the binomial series formula?

Usually you will be asked to expand something in the form ${(p + q x)}^{n}$
But the formula only works if the constant term is a 1
- So start by pulling out a factor of $p$
  - ${(p + q x)}^{n} = {(p (1 + \frac{q}{p} x))}^{n} = p^{n} {(1 + \frac{q}{p} x)}^{n}$
- Then expand ${(1 + \frac{q}{p} x)}^{n}$
  - Substitute $\frac{q}{p} x$ everywhere that $x$ is in the formula
  - The interval of convergence becomes $|\frac{p}{q} x| < 1$
- Don't forget to multiply everything by $p^{n}$ again at the end!
Be sure you can recognise a negative or fractional power
- The expression may be in the denominator of a fraction
  - $\frac{1}{{(p + q x)}^{k}} = {(p + q x)}^{- k}$
- Or inside a square root
  - $\sqrt{p + q x} = {(p + q x)}^{\frac{1}{2}}$
- Or be written as a more complex root
  - $\sqrt[m]{{(p + q x)}^{k}} = {(p + q x)}^{\frac{k}{m}}$

Examiner Tips and Tricks

Remember the formula is on the formula sheet
Be especially careful with
- negative numbers
- subtracting 1 from fractions
Use brackets to separate things out
Don't rush!

Worked Example

(a) Expand $\frac{1}{\sqrt{9 - 3 x}}$ in ascending powers of $x$ up to and including the term in $x^{3}$ and simplifying each term as far as possible.

Start by rewriting using laws of indices

$\frac{1}{\sqrt{9 - 3 x}} = {(9 - 3 x)}^{- \frac{1}{2}}$

Now pull out a factor to make the constant term inside the brackets a 1

$\begin{array}{rcl} {(9 - 3 x)}^{- \frac{1}{2}} & = & (9^{- \frac{1}{2}}) {(1 - \frac{3}{9} x)}^{- \frac{1}{2}} \\ = & \frac{1}{3} {(1 - \frac{x}{3})}^{- \frac{1}{2}} \end{array}$

Now use the binomial series formula to expand ${(1 - \frac{x}{3})}^{- \frac{1}{2}}$

Use $n = \frac{1}{2}$ and substitute $- \frac{x}{3}$ everywhere that $x$ appears in the formula

$\begin{array}{rcl} {(1 - \frac{x}{3})}^{- \frac{1}{2}} & = & 1 + (- \frac{1}{2}) (- \frac{x}{3}) + \frac{(- \frac{1}{2}) (- \frac{1}{2} - 1)}{2!} {(- \frac{x}{3})}^{2} + \frac{(- \frac{1}{2}) (- \frac{1}{2} - 1) (- \frac{1}{2} - 2)}{3!} {(- \frac{x}{3})}^{3} + . . . \\ = & 1 + \frac{1}{6} x + \frac{(- \frac{1}{2}) (- \frac{3}{2})}{2} (\frac{x^{2}}{9}) + \frac{(- \frac{1}{2}) (- \frac{3}{2}) (- \frac{5}{2})}{6} (- \frac{x^{3}}{27}) + . . . \\ = & 1 + \frac{1}{6} x + \frac{(\frac{3}{4})}{2} (\frac{x^{2}}{9}) + \frac{(- \frac{15}{8})}{6} (- \frac{x^{3}}{27}) + . . . \\ = & 1 + \frac{1}{6} x + (\frac{3}{8}) (\frac{x^{2}}{9}) + (- \frac{5}{16}) (- \frac{x^{3}}{27}) + . . . \\ = & 1 + \frac{1}{6} x + \frac{1}{24} x^{2} + \frac{5}{432} x^{3} + . . . \end{array}$

Now don't forget to multiply by $\frac{1}{3}$ (factorised out earlier) to get the final answer!

$\frac{1}{\sqrt{9 - 3 x}} = \frac{1}{3} (\begin{array}{rcl} 1 + \frac{1}{6} x + \frac{1}{24} x^{2} + \frac{5}{432} x^{3} + . . . \end{array})$

$\frac{1}{\sqrt{9 - 3 x}} = \frac{1}{3} + \frac{1}{18} x + \frac{1}{72} x^{2} + \frac{5}{1296} x^{3} + . . .$

(b) Find the interval of convergence for the expansion in part (a).

Remember that we used $- \frac{x}{3}$ in place of $x$ when we used the binomial series formula

We also need to substitute $- \frac{x}{3}$ into the standard convergence interval $|x| < 1$

$|- \frac{x}{3}| < 1 \frac{1}{3} |x| < 1 - 1 < \frac{1}{3} x < 1$

$- 3 < x < 3$

Last updated: 20 October 2024

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