Binomial Expansion - The Essential Guide For Year 13 Students
Introduction
When it comes to the binomial expansion and A Level Maths you will have met it in year 1 and you will also meet it in year 2.
In year 1, you will follow Pascal’s Triangle and/or the NCR method. Both of them rely on positive powers.
In year 2, you will look at expansions where the power is either negative or a fractional power. It is important to note that these questions can be quite algebraic intensive especially if you are dealing with questions that also involve partial fractions.
The Binomial Expansion Formula
Here is the formula which we will be using when going over some examples.
The formula is:
(1+x)^n=1+n x+\frac{n(n-1)}{2 !} x^2+\frac{n(n-1)(n-2)}{3 !} x^3+\ldotsWhat is important is that what you have inside the brackets must begin with a “1”. We will see what happens if there is something other than one in a later example.
First you need to rewrite the expansion in terms of a fractional power. Then you can use the formula to find the expansion. Note that the question is asking you not to go beyond x^3 so it is important that you do not.
\begin{aligned} (1+x)^{\frac{1}{2}} & =1+\frac{1}{2}(x)+\frac{\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)}{2 !}(x)^2+\frac{\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{3 !}(x)^3 \ldots \\ & =1+\frac{1}{2} x-\frac{1}{8} x^2+\frac{1}{16} x^3 \cdots \end{aligned}
In terms of what the expansion is valid for this is |x|<1
Binomial Expansion - What to do when you don't see a “1”?
Let us now consider the following example:
Here you can clearly see that you have a number 4 as opposed to the number 1. In this instance, the formula cannot be used. Before you can use the formula you need to first of all take out a factor. In this case the factor is 4.
\begin{aligned}
(4+x)^{\frac{1}{2}} & =\left[4\left(1+\frac{x}{4}\right)\right]^{\frac{1}{2}} \\
& =2\left(1+\frac{x}{4}\right)^{\frac{1}{2}}
\end{aligned}
You can see how the factor of 4 has been taken out. What you also need to remember is that any factor that you take out is still being raised to the power of the original expansion. You can see that there is now a binomial expansion that contains a “1” which can now be expanded and this is then multiplied by the 2.
The whole solution is presented below:
\begin{aligned} & (4+x)^{\frac{1}{2}}=\left[4\left(1+\frac{x}{4}\right)\right]^{\frac{1}{2}} \\ & =2\left(1+\frac{x}{4}\right)^{\frac{1}{2}} \\ & =2\left[1+\frac{1}{2}\left(\frac{x}{4}\right)+\frac{\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)}{2 !}\left(\frac{x}{4}\right)^2+\frac{\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{3 !}\left(\frac{x}{4}\right)^3\right] \\ & =2\left[1+\frac{x}{8}-\frac{x^2}{128}+\frac{x^3}{1024} \cdots\right]=2+\frac{x}{4}-\frac{x^2}{64}+\frac{x^2}{512} \cdots \\ & \end{aligned}The expansion is valid for \left|\frac{x}{4}\right|<1 \text { ie }|x|<4
Binomial Expansion – When there are two expansions
Let us consider the above example. In this instance there will be two expansions that need to be created and again you need to read the question in terms of how many terms are required.
First you need to rearrange the expansion so it is in terms of a fractional power as follows:
\frac{(1+2 x)^{\frac{1}{2}}}{(1-x)^{\frac{1}{2}}}And the above is the same as:
(1+2 x)^{\frac{1}{2}}(1-x)^{-\frac{1}{2}}\begin{aligned} & (1+2 x)^{\frac{1}{2}}=1+\frac{1}{2}(2 x)+\frac{\frac{1}{2}\left(-\frac{1}{2}\right)}{2 !}(2 x)^2 \\ & =1+x-\frac{1}{2} x^2 \ldots \quad<\quad \text { valid }|2 x|<1 \text { ie }|x|<\frac{1}{2} \end{aligned}\begin{aligned} (1-x)^{-\frac{1}{2}} & =1+\left(-\frac{1}{2}\right)(-1)+\frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{2 !}(-x)^2 \ldots \\ & =1+\frac{1}{2} x+\frac{3}{8} x^2+\cdots \end{aligned}The expansion is valid for |-x|<1 \text { ie }|x|<1
Putting the expansions together and collecting like terms gives:
\begin{aligned} =\left(1+x-\frac{1}{2} x^2 \cdots\right)\left(1+\frac{1}{2} x+\frac{3}{8} x^2 \cdots\right) & \\ = & +\frac{1}{2} x+\frac{3}{8} x^2 \\ & +x+\frac{1}{2} x^2 \\ & -\frac{1}{2} x^2 \\ =1 & +\frac{3}{2} x+\frac{3}{8} x^2 \ldots \end{aligned}You can see that there have been two expansions. In order to determine what the expansion is valid for, always pick the stricter restriction. So overall, the expansion is valid for |x|<\frac{1}{2}
Binomial Expansion – Partial Fractions
Note that in order to do this question you first need to write the algebraic fraction that you have been given as partial fractions. The question is not asking you to do this, you need to be able to spot that this is indeed the method that you have to follow.
For the purpose of this article, it will be taken that you understand how to do partial fractions and that you will know how to obtain the answer of \frac{1}{x-1}+\frac{1}{2-x}
There are then two binomial expansions that need to be performed: =(x-1)^{-1}+(2-x)^{-1}
Dealing with each in turn we will have the following:
\begin{aligned} (x-1)^{-1}=(-1+x)^{-1} & =[-1(1-x)]^{-1} \\ & =-1(1-x)^{-1} \\ & =-1\left[1+(-1)(-x)+\frac{(-1)(-2)}{2 !}(-x)^2+\frac{(-1)(-2)(-3)}{3 !}(-x)^3\right] \\ & =-1\left[1+x+x^2+x^3\right] \\ & =-1-x-x^2-x^3 \end{aligned}\begin{aligned} (2-x)^{-1} & =\left[2\left(1-\frac{x}{2}\right)\right]^{-1}=2^{-1}\left(1-\frac{x}{2}\right)^{-1} \\ & =\frac{1}{2}\left[1+(-1)\left(-\frac{x}{2}\right)+\frac{(-1)(-2)}{2 !}\left(-\frac{x}{2}\right)^2+\frac{(-1)(-2)(-3)}{3 !}\left(-\frac{x}{2}\right)^3\right] \\ & =\frac{1}{2}\left[1+\frac{x}{2}+\frac{x^2}{4}+\frac{x^3}{8}\right] \\ & =\frac{1}{2}+\frac{x}{4}+\frac{x^2}{8}+\frac{x^3}{16} \end{aligned}Adding the two expansions gives -\frac{1}{2}-\frac{3}{4} x-\frac{7}{8} x^2-\frac{15}{16} x^3
The overall maths behind binomial expansion is not difficult, but there is a lot of algebra and so it is important that you are giving yourself plenty of room to work and that your writing is legible. With all the algebra and powers, it would be very easy to make a mistake.