Examples
Pascal's triangle, then, makes it easy to perform binomial expansions, allowing you to just write the terms straight down without any of the awkward bracket multiplications. There are however, two limitations to this method.
We therefore need to find a formula that will tell us the coefficients of an expansion directly.
You do not need to know why the binomial formula works, but it may prove interesting. You may, however,
It was explained earlier that each term in a binomial expansion is formed by taking one of the terms from each bracket. To see what this means, consider the simple example
Let's look at one more example
How does this help us find a formula?
In the above case the co-efficient of a2b was 3. This was because there are 3 ways of going through 3 brackets and choosing 2 a's and 1 b. That is all the co-efficient tells you: "How many ways can you go through all the brackets, picking up so many a's and so many b's?"
So, for the expansion of (a+b)9, the coefficient of the a4b5 term will just be the number of ways of choosing 4 a's and 5 b's from 9 brackets. This is just the theory of combinations, a central part of probability.
Combinations
Let's consider the example above. How many ways can you go through 9 brackets and collect 4 a's?
For the first a we have 9 brackets to choose from, so there are 9 ways.
For the second, we have already used one bracket, so there are now 8 ways, and so on.
There is, however, a problem with this reasoning. If I label the bracket that an a comes from with a subscript,( e.g. the a from the seventh bracket is a7), then we can see that there is more than one way of choosing the same a's.
If I chose my a's from the first 4 brackets, in the correct order, I would get:
a1a2a3a4
I could, however, have chosen the second bracket first and then the first bracket, giving
a2a1a3a4
However, these are clearly the same result. I have therefore counted this result too many times.
To work out how many times I have over-counted, I need to find how many different ways I could arrange the a's. I have 4 choices for the first a, but then only 3 left for the second choice, then 2 for the third choice and 1 for the final choice. Combining these, there are 4x3x2x1 choices. When I worked out the co-efficient I was 24 times too large!
The General Binomial Theorem
The expansion can therefore be written
To check that this formula really works, let's try a simple example.
So the formula works!
In fact the formula will work for any real value of n( fractions and negatives included).
Examples of how the formula can be used are on the next page.
Reminder, the binomial formula is
When you have more complicated terms inside the brackets, you must be very careful about how you write the first line of working.
You may be asked to find all the terms up to a certain power of x. You must be quite sure that you have fulfilled the requirement before you stop.
You will need to be a bit more careful if powers or fractions are included in the brackets.
One application of the binomial theorem is easily approximating expressions like (0.98)10, which will be demonstrated on the next page.
Example 1
What are the first 5 terms in the expansion of
We can use this expansion to make a quick approximation of (0.98)10.
All the x terms are doing are moving the decimal points in the terms, so the sum is a very easy one.
Example 2
Find, to 3 d.p., the value of (1.03)7.
