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 \(a^2 + 2ab + b^2\) 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
\[(a + b)^n = \sum_{r=0}^n \binom{n}{r} a^rb^{n-r}\] 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.
\[\begin{align} (x+2)^4 &= \binom{4}{0} x^0\times 2^4 + \binom{4}{1} x^1\times 2^3 + \binom{4}{2} x^2\times 2^2 + \binom{4}{3} x^3\times 2^1 + \binom{4}{4} x^4\times 2^0 \\ &= 1 \times 1 \times 16 + 4 \times x \times 8 + 6 \times x^2 \times 4 + 4 \times x^3 \times 2 + 1 \times x^4 \times 1\\ &= 16 + 32x + 24x^2 + 8x^3 + x^4 \end{align}\]
Where people often make mistakes is when the terms in the binomial are a bit more complicated, e.g. \((2x -3)\). It is vitally important to use brackets when doing the expansion in this case. E.g.
\[\begin{align} (2x-3)^4 &= \binom{4}{0} (2x)^0\times (-3)^4 + \binom{4}{1} (2x)^1\times (-3)^3 + \binom{4}{2} (2x)^2\times (-3)^2 + \binom{4}{3} (2x)^3\times (-3)^1 \\ & \qquad + \binom{4}{4} (2x)^4\times (-3)^0 \\ &= 1 \times 1 \times 81 + 4 \times 2x \times (-27) + 6 \times 4x^2 \times 9 + 4 \times 8x^3 \times (-3) + 1 \times 16x^4 \times 1\\ &= 81 - 54x + 36x^2 - 24x^3 + 16x^4 \end{align}\]
7.2 The expansion for all powers
If the index, n, is not a positive integer, e.g. \(n=-2\) or \(n = {1 \over 2}\), the expansion given in the last section can’t be used, and the correct expansion is:
\[\begin{equation} (1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)...(n-r+1)}{r!}x^r + ... \qquad (|x| < 1, n \in \mathbb{R} ) \tag{7.1} \end{equation}\]
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
\[\begin{align} (1+x)^3 &= 1 + (-2)x + \frac{(-2)(-2-1)}{2!}x^2 + \frac{(-2)(-2-1)(-2-2)}{3!}x^3 + ... \qquad (|x| < 1)\\ &= 1 + (-2)x + \frac{(-2)(-3)}{2\times 1}x^2 + \frac{(-2)(-3)(-4)}{3\times 2 \times 1}x^3 + ...\\ &= 1 -2x + 3x^2 - 4x^3 + ... \end{align}\]
Just as in the last case, brackets are your friends when you have more complicated terms in the binomial.
E.g. Expanding \((1-2x)^{-2}\) up to and including terms in \(x^3\), would give
\[\begin{align} (1-2x)^3 &= 1 + (-2)(-2x) + \frac{(-2)(-2-1)}{2!}(-2x)^2 + \frac{(-2)(-2-1)(-2-2)}{3!}(-2x)^3 + ... \qquad (|x| < 1)\\ &= 1 + (-2)(-2)x + \frac{(-2)(-3)}{2\times 1}(4)x^2 + \frac{(-2)(-3)(-4)}{3\times 2 \times 1}(-8)x^3 + ...\\ &= 1 +4x + 12x^2 + 32x^3 + ... \end{align}\]
Note that for this expansion we need \(|-2x| < 1\) which means \(|x| < {1 \over 2}\).
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 \[ (3+x)^{-2} = \left(3\left(1+\frac{x}{3}\right)\right)^{-2}\]
- Take the 3 out of the outer set of brackets \[ (3+x)^{-2} = (3)^{-2}\left(1+\frac{x}{3}\right)^{-2}\]
- This has left us with \(\left(1+\frac{x}{3}\right)^{-2}\) which is in the standard form, and can be expanded as before \[\begin{align} (3+x)^{-2} &= (3)^{-2}\left(1+(-2)(\frac{x}{3}) + \frac{(-2)(-3)}{2!}(\frac{x}{3})^2+...\right) \\ &= \frac{1}{9}\left(1-\frac{2}{3}x + \frac{1}{3}x^2+...\right) \\ &= \frac{1}{9}-\frac{2}{27}x + \frac{1}{27}x^2+... \end{align}\]