Topic 7 The Binomial Expansion

You should be very familiar with expanding expressions such as $(a + b)^{2}$ . You may even be happy to immediately write down the expansion as $a^{2} + 2 a b + b^{2}$ without any intermediate working. This is an example of a binomial expansion (binomial just means two numbers). In general we might want to expand $(a + b)^{n}$ where n could be any number.

There are two cases:

n is a positive integer. In this case you could just write it as n seperate brackets and expand each one in turn; this would work, but is tedious. The next section details this case.
n is not a positive integer. In this case there are limits on how you can use the expansion. There is a separate section on this.

7.1 The expansion for positive integer powers

For any non-negative integer $n$ , we have

$(a + b)^{n} = \sum_{r = 0}^{n} (\binom{n}{r}) a^{r} b^{n - r}$ where

$(\binom{n}{r}) = \frac{n!}{r! (n - r)!}$

is the number of ways of choosing r objects out of n objects if we don’t care about the order (and can be read as “n choose r”). Fortunately most scientific calculators have a button, normally labelled “nCr” which calculates this for you.

E.g.

$\begin{aligned} (x + 2)^{4} & = (\binom{4}{0}) x^{0} \times 2^{4} + (\binom{4}{1}) x^{1} \times 2^{3} + (\binom{4}{2}) x^{2} \times 2^{2} + (\binom{4}{3}) x^{3} \times 2^{1} + (\binom{4}{4}) x^{4} \times 2^{0} \\ = 1 \times 1 \times 16 + 4 \times x \times 8 + 6 \times x^{2} \times 4 + 4 \times x^{3} \times 2 + 1 \times x^{4} \times 1 \\ = 16 + 32 x + 24 x^{2} + 8 x^{3} + x^{4} \end{aligned}$

Where people often make mistakes is when the terms in the binomial are a bit more complicated, e.g. $(2 x - 3)$ . It is vitally important to use brackets when doing the expansion in this case. E.g.

$\begin{aligned} (2 x - 3)^{4} & = (\binom{4}{0}) (2 x)^{0} \times (- 3)^{4} + (\binom{4}{1}) (2 x)^{1} \times (- 3)^{3} + (\binom{4}{2}) (2 x)^{2} \times (- 3)^{2} + (\binom{4}{3}) (2 x)^{3} \times (- 3)^{1} \\ + (\binom{4}{4}) (2 x)^{4} \times (- 3)^{0} \\ = 1 \times 1 \times 81 + 4 \times 2 x \times (- 27) + 6 \times 4 x^{2} \times 9 + 4 \times 8 x^{3} \times (- 3) + 1 \times 16 x^{4} \times 1 \\ = 81 - 54 x + 36 x^{2} - 24 x^{3} + 16 x^{4} \end{aligned}$

7.2 The expansion for all powers

If the index, n, is not a positive integer, e.g. $n = - 2$ or $n = \frac{1}{2}$ , the expansion given in the last section can’t be used, and the correct expansion is:

$\begin{matrix} (7.1) & (1 + x)^{n} = 1 + n x + \frac{n (n - 1)}{2!} x^{2} + \frac{n (n - 1) . . . (n - r + 1)}{r!} x^{r} + . . . (| x | < 1, n \in R) \end{matrix}$

Note that this expansion won’t end, and is only valid if x is small enough.

E.g. Expanding $(1 + x)^{- 2}$ up to and including terms in $x^{3}$ , would give

$\begin{aligned} (1 + x)^{3} & = 1 + (- 2) x + \frac{(- 2) (- 2 - 1)}{2!} x^{2} + \frac{(- 2) (- 2 - 1) (- 2 - 2)}{3!} x^{3} + . . . (| x | < 1) \\ = 1 + (- 2) x + \frac{(- 2) (- 3)}{2 \times 1} x^{2} + \frac{(- 2) (- 3) (- 4)}{3 \times 2 \times 1} x^{3} + . . . \\ = 1 - 2 x + 3 x^{2} - 4 x^{3} + . . . \end{aligned}$

Just as in the last case, brackets are your friends when you have more complicated terms in the binomial.

E.g. Expanding $(1 - 2 x)^{- 2}$ up to and including terms in $x^{3}$ , would give

$\begin{aligned} (1 - 2 x)^{3} & = 1 + (- 2) (- 2 x) + \frac{(- 2) (- 2 - 1)}{2!} (- 2 x)^{2} + \frac{(- 2) (- 2 - 1) (- 2 - 2)}{3!} (- 2 x)^{3} + . . . (| x | < 1) \\ = 1 + (- 2) (- 2) x + \frac{(- 2) (- 3)}{2 \times 1} (4) x^{2} + \frac{(- 2) (- 3) (- 4)}{3 \times 2 \times 1} (- 8) x^{3} + . . . \\ = 1 + 4 x + 12 x^{2} + 32 x^{3} + . . . \end{aligned}$

Note that for this expansion we need $| - 2 x | < 1$ which means $| x | < \frac{1}{2}$ .

In the previous section we looked at $(a + b)^{n}$ , whereas in this section we have looked at $(1 + x)^{n}$ . If we want to use equation (7.1) for an expression like $(3 + x)^{- 2}$ it has to be transformed into something which looks like $(1 + x)^{n}$ and then expanded as normal.

E.g.

Divide the expression by 3 $(3 + x)^{- 2} = {(3 (1 + \frac{x}{3}))}^{- 2}$
Take the 3 out of the outer set of brackets $(3 + x)^{- 2} = (3)^{- 2} {(1 + \frac{x}{3})}^{- 2}$
This has left us with ${(1 + \frac{x}{3})}^{- 2}$ which is in the standard form, and can be expanded as before $\begin{aligned} (3 + x)^{- 2} & = (3)^{- 2} (1 + (- 2) (\frac{x}{3}) + \frac{(- 2) (- 3)}{2!} (\frac{x}{3})^{2} + . . .) \\ = \frac{1}{9} (1 - \frac{2}{3} x + \frac{1}{3} x^{2} + . . .) \\ = \frac{1}{9} - \frac{2}{27} x + \frac{1}{27} x^{2} + . . . \end{aligned}$

LJMU Maths Refresher

Topic 7 The Binomial Expansion

7.1 The expansion for positive integer powers

7.2 The expansion for all powers

7.3 Further Resources