General Binomial Expansion

- If 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 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

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

(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

General Binomial Expansion (Edexcel IGCSE Further Maths)