M1 Study Notes: Binomial Expansion
Hello Future Maths Whiz!
Ever looked at something like $$(a+b)^7$$ and thought, "There has to be a better way than multiplying this out seven times!"? Well, you're in luck! This chapter is all about that "better way."
We're going to learn about the Binomial Expansion, a powerful shortcut for expanding expressions that have two terms (a "bi-nomial") raised to a power. It's a super important tool in M1 that saves a ton of time and pops up in other areas like probability. Let's get started!
1. The Pattern Behind the Power
Let's start by doing it the long way for small powers and see if we can spot a pattern.
$$(a+b)^0 = 1$$
$$(a+b)^1 = 1a + 1b$$
$$(a+b)^2 = 1a^2 + 2ab + 1b^2$$
$$(a+b)^3 = 1a^3 + 3a^2b + 3ab^2 + 1b^3$$
$$(a+b)^4 = 1a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + 1b^4$$
Do you see what's happening?
- Powers of 'a': They start at the highest power (n) and decrease by 1 in each term, all the way down to 0.
- Powers of 'b': They do the opposite! They start at 0 and increase by 1 in each term, up to the highest power (n).
- The 'Coefficients': These are the numbers in front of the terms. Notice the symmetrical pattern: 1, 4, 6, 4, 1. But where do these numbers come from?
Introducing Pascal's Triangle
A French mathematician named Blaise Pascal noticed a beautiful pattern in these coefficients, which we now call Pascal's Triangle. It's a simple, visual way to find the coefficients for your expansion.
How to build it:
1. Start with a '1' at the top.
2. Each new row starts and ends with a '1'.
3. Every other number is found by adding the two numbers directly above it.
It looks like this:
Row 0 (for n=0): 1
Row 1 (for n=1): 1 1
Row 2 (for n=2): 1 2 1
Row 3 (for n=3): 1 3 3 1
Row 4 (for n=4): 1 4 6 4 1
Row 5 (for n=5): 1 5 10 10 5 1
Example: Expand $$(x+y)^4$$ using Pascal's Triangle
1. Find the coefficients: The power is 4, so we look at Row 4 of the triangle. The coefficients are 1, 4, 6, 4, 1.
2. Add the variables: The powers of *x* will go down from 4 to 0, and the powers of *y* will go up from 0 to 4.
3. Combine them:
$$(x+y)^4 = \mathbf{1}x^4y^0 + \mathbf{4}x^3y^1 + \mathbf{6}x^2y^2 + \mathbf{4}x^1y^3 + \mathbf{1}x^0y^4$$
Cleaning it up gives:
$$(x+y)^4 = x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + y^4$$
Key Takeaway
Pascal's Triangle is a quick and easy way to find the coefficients for binomial expansions, especially for smaller powers.
2. A More Powerful Method: Combinations
Pascal's Triangle is great, but what if you need to expand something to the power of 20? You don't want to write out 20 rows of the triangle! We need a more direct formula. This is where Combinations come in.
Quick Review: What are Combinations?
Remember from your junior form stats that $$C^n_r$$ (or $$\binom{n}{r}$$) means "the number of ways to choose *r* items from a set of *n* items".
The formula is $$C^n_r = \frac{n!}{r!(n-r)!}$$, but the best part is that your calculator has an nCr button! Use it to save time.
For example, to find the '6' in the middle of Row 4 of Pascal's Triangle, we can calculate $$C^4_2$$. Try it on your calculator!
The coefficients in Row *n* of Pascal's Triangle are just:
$$C^n_0, C^n_1, C^n_2, C^n_3, ..., C^n_n$$
The Binomial Theorem: The Master Formula
Now we can put everything together into one awesome formula called the Binomial Theorem.
For any positive integer *n*:
$$(a+b)^n = C^n_0a^n + C^n_1a^{n-1}b^1 + C^n_2a^{n-2}b^2 + ... + C^n_ra^{n-r}b^r + ... + C^n_na^0b^n$$
Memory Aid: The Power Rule!
For any term in the expansion, the power of *a* plus the power of *b* always adds up to *n*. For the term $$C^n_r a^{n-r}b^r$$, the powers are $$(n-r) + r = n$$. Easy!
Step-by-Step Example: Expand $$(2x - 1)^4$$
Don't worry, we'll take this one step at a time.
Step 1: Identify your pieces.
In $$(a+b)^n$$, we have:
- n = 4
- a = 2x
- b = -1 (Super important! Always take the sign with the term!)
Step 2: Write out the expansion using the formula as a template.
$$(2x-1)^4 = C^4_0(2x)^4(-1)^0 + C^4_1(2x)^3(-1)^1 + C^4_2(2x)^2(-1)^2 + C^4_3(2x)^1(-1)^3 + C^4_4(2x)^0(-1)^4$$
Step 3: Calculate the coefficients ($$C^n_r$$ values).
$$C^4_0=1, C^4_1=4, C^4_2=6, C^4_3=4, C^4_4=1$$
Step 4: Substitute the coefficients and simplify each term carefully.
- Term 1: $$1 \cdot (16x^4) \cdot (1) = 16x^4$$
- Term 2: $$4 \cdot (8x^3) \cdot (-1) = -32x^3$$
- Term 3: $$6 \cdot (4x^2) \cdot (1) = 24x^2$$
- Term 4: $$4 \cdot (2x) \cdot (-1) = -8x$$
- Term 5: $$1 \cdot (1) \cdot (1) = 1$$
Step 5: Put it all together.
$$(2x-1)^4 = 16x^4 - 32x^3 + 24x^2 - 8x + 1$$
Common Mistakes to Avoid!
- The Negative Sign: Forgetting to include the negative sign with the 'b' term (like in $$(a-b)^n$$) is the most common error. Remember, $$b$$ is negative!
- Powering the Whole Term: When you have something like $$(2x)^3$$, remember to cube BOTH the 2 and the x. It's $$8x^3$$, not $$2x^3$$.
3. Finding Just One Specific Term
Sometimes a question won't ask for the full expansion. It might ask for just the 5th term, or the term containing $$x^7$$. For this, we use the General Term formula.
The formula for any term in the expansion is:
General Term: $$T_{r+1} = C^n_r a^{n-r} b^r$$
Wait, why is it $$T_{r+1}$$? This is a little tricky, but it's because the first term uses r=0 ($$C^n_0$$), the second term uses r=1 ($$C^n_1$$), and so on. So the term number is always one more than the value of *r*.
- To find the 1st term ($$T_1$$), use r = 0.
- To find the 3rd term ($$T_3$$), use r = 2.
- To find the 10th term ($$T_{10}$$), use r = 9.
Example 1: Find the 4th term of $$(x+2y)^9$$
1. Identify the pieces: n=9, a=x, b=2y.
2. Find r: For the 4th term ($$T_4$$), we use r = 3.
3. Plug into the general term formula:
$$T_{3+1} = C^9_3 (x)^{9-3} (2y)^3$$
$$T_4 = 84 \cdot x^6 \cdot (8y^3)$$
$$T_4 = 672x^6y^3$$
Example 2: Find the term containing $$x^8$$ in the expansion of $$(x^2 + \frac{3}{x})^7$$
This is a classic exam-style question! Don't panic.
1. Identify the pieces: n=7, a=$$x^2$$, b=$$3x^{-1}$$. (It's often easier to write fractions like $$\frac{3}{x}$$ with negative powers).
2. Write the general term, but leave 'r' as a variable for now.
$$T_{r+1} = C^7_r (x^2)^{7-r} (3x^{-1})^r$$
3. Focus ONLY on the powers of x and simplify them.
$$T_{r+1} = C^7_r \cdot 3^r \cdot (x^2)^{7-r} \cdot (x^{-1})^r$$
$$T_{r+1} = ... \cdot x^{2(7-r)} \cdot x^{-r}$$
$$T_{r+1} = ... \cdot x^{14-2r-r}$$
$$T_{r+1} = ... \cdot x^{14-3r}$$
4. Set the power of x equal to what you need (in this case, 8) and solve for r.
$$14 - 3r = 8$$
$$6 = 3r$$
$$r = 2$$
5. Now you know r=2! Plug it back into the full general term formula to find the complete term.
$$T_{2+1} = C^7_2 (x^2)^{7-2} (3x^{-1})^2$$
$$T_3 = 21 \cdot (x^2)^5 \cdot (3^2(x^{-2}))$$
$$T_3 = 21 \cdot x^{10} \cdot 9x^{-2}$$
$$T_3 = 189 x^8$$
So, the term containing $$x^8$$ is $$189x^8$$.
Key Takeaway
The general term formula $$T_{r+1} = C^n_r a^{n-r} b^r$$ is your best friend for finding specific terms. Just remember that the term number is always `r+1`.
4. A Final Note: Summation Notation ($$\Sigma$$)
As per the syllabus, you need to recognize the binomial expansion written in a compact form using the Greek letter Sigma ($$\Sigma$$), which means "sum up".
The full Binomial Theorem:
$$(a+b)^n = C^n_0a^nb^0 + C^n_1a^{n-1}b^1 + ... + C^n_na^0b^n$$
Can be written as:
$$(a+b)^n = \sum_{r=0}^{n} C^n_r a^{n-r} b^r$$
This notation simply means: "Start with r=0, plug it into the formula $$C^n_r a^{n-r} b^r$$. Then, do it again for r=1, then r=2, and so on, all the way up to r=n. Finally, add all of those results together."
You don't need to do complex calculations with it, but you should recognize that it's just a shorthand way of writing the Binomial Theorem.
Chapter Summary
Quick Review Box
- The Binomial Theorem:
$$(a+b)^n = \sum_{r=0}^{n} C^n_r a^{n-r} b^r$$ - The General Term (for the $$(r+1)$$-th term):
$$T_{r+1} = C^n_r a^{n-r} b^r$$
We've learned two main ways to find the coefficients: Pascal's Triangle (great for small *n*) and Combinations $$C^n_r$$ (works for everything!). Remember the key patterns: powers of 'a' go down, powers of 'b' go up, and they always sum to *n*.
This is a foundational topic, so make sure you practice a few expansions on your own. You've got this!