General Binomial Expansion (Edexcel IGCSE Further Maths)

Spec point — 6A

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$n\inℚ$

    • $ℚ$ 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 $n\inℚ$ 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

  • $( 1 + x )^{n} = 1 + n x + \frac{n ( n - 1 )}{2 factorial} x^{2} +... + \frac{n ( n - 1 )... ( n - r + 1 )}{r factorial} x^{r} +... \text{for} \left| x \right| < 1 , n \in \mathbb{Q}$

  • 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 open vertical bar x close vertical bar less than 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 open vertical bar p over q x close vertical bar less than 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

    • $\sqrt{p + q x} = ( p + q x )^{1 \frac{1}{2}}$

  • Or be written as a more complex root

    • $\sqrt[m]{ ( p + q x ) ^{k}} = ( p + q x ) ^{\frac{k}{m}}$

Exam Hint

• 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 $\frac{1}{\sqrt{9-3x}}$ 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

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

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

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

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

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

| ( 1 - x / 3 ) ^ - 1 half | = 1 + ( - 1 half ) ( - x / 3 ) + (( - 1 half ) ( - 1 half - 1 ))/(2 factorial) ( - x / 3 ) ^2 + (( - 1 half ) ( - 1 half - 1 ) ( - 1 half - 2 ))/(3 factorial) ( - x / 3 ) ^3 +... | | _ = 1 + 1 / 6 x + (( - 1 half ) ( - 3 / 2 ))/(2) ( x ^2 / 9 ) + (( - 1 half ) ( - 3 / 2 ) ( - 5 / 2 ))/(6) ( - x ^3 / 27 ) +... | | _ = 1 + 1 / 6 x + (( 3 / 4 ))/(2) ( x ^2 / 9 ) + (( - 15 / 8 ))/(6) ( - x ^3 / 27 ) +... | | _ = 1 + 1 / 6 x + ( 3 / 8 ) ( x ^2 / 9 ) + ( - 5 / 16 ) ( - x ^3 / 27 ) +... | | _ = 1 + 1 / 6 x + 1 / 24 x ^2 + 5 / 432 x ^3 +... | |

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

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

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

(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 open vertical bar x close vertical bar less than 1

open vertical bar negative x over 3 close vertical bar less than 1 1 third open vertical bar x close vertical bar less than 1 minus 1 less than 1 third x less than 1

$-3<x<3$