The Binomial Theorem
- The binomial theorem is a method for expanding powers of binomials.
- For any positive integer $n$, it gives the expansion of $(a+b)^n$ as a sum of terms involving powers of $a$ and $b$.
Pascal's Triangle and nCr Notation
Pascal's Triangle
- Pascal's triangle is a triangular array of the binomial coefficients that arises in probability theory, combinatorics, and algebra.
- Each number is the sum of the two numbers directly above it.
Note
- The numbers in Pascal's triangle correspond to the coefficients in the binomial expansion.
- The nth row (starting with row 0 at the top) gives the coefficients of the expansion of $(a+b)^n$.
Hint
- It's worth it memorizing the first few rows of Pascal's triangle, up to the 6th row (1, 5, 10, 10, 5, 1).
- This will help you expand any binomial power up to $(a + b)^5$.
Note
We'll talk about Pascal's triangle later in the section on calculating binomial coefficients.
nCr Notation
- The notation $^nC_r$ (or $\binom{n}{r}$) represents the number of ways to choose $r$ items from a set of $n$ items, where order doesn't matter.
- It's read as "n choose r".
- The formula for the number of combinations is calculated as: $$ nCr = \frac{n!}{r!(n-r)!} $$
Example
- For example, let's take a set of four items, A, B, C, and D.
- If we want to choose two items from this set, these are the possible combinations:
- A, B
- A, C
- A, D
- B, C
- B, D
- C, D
- There are six possible combinations, so we say that $^4C_2 = 6$.
Note
- There are many alternative notations for $^nC_r$, such as $^n_rC$, $nCr$, $C^n_r$, and $\binom{n}{r}$.
- All of these are interchangeable, so use whichever one you prefer.
Example question
Calculate $^5C_2$.
Solution
$$^5C_2 = \frac{5!}{2!(5-2)!} = \frac{5 \cdot 4}{2 \cdot 1} = 10$$
This means there are 10 ways to choose 2 items from a set of 5.
Hint
When calculating $^nC_r$, it's often easier to cancel out common factors in the numerator and denominator before multiplying.
$^nC_r$ also corresponds to Pascal's triangle, and is equal to the $r$th number on the $n$th row, starting counting from $0$.
Common Mistake
- If you use this method, don't forget to start counting from $0$.
- The first row of the triangle is $0$, and the column of $1$s at the left is also $0$. (PS If you take CS, this should be natural to you)
Calculating $^nC_r$
Using the Formula
As mentioned earlier, $^nC_r$ can be calculated using the formula:
$$ ^nC_r = \frac{n!}{r!(n-r)!} $$
Using Technology
- Many calculators have a built-in $^nC_r$ function.
- On a graphing calculator, you can often find this in the MATH or PROB (probability) menu.
Hint
When $n$ and $r$ are large, using technology is often faster and less prone to calculation errors than using the formula directly.
Hint
- Due to the symmetry of Pascal's triangle, $^nC_r = ^nC_{(n-r)}$.
- This is why the values in the table above are symmetric.
Expansion of $(a+b)^n$
- The binomial theorem states that for any positive integer $n$: $$(a+b)^n = \sum_{k=0}^n \binom{n}{k} a^{n-k} b^k$$
- This can be written out as: $$(a+b)^n = \binom{n}{0}a^n + \binom{n}{1}a^{n-1}b + \binom{n}{2}a^{n-2}b^2 + ... + \binom{n}{n-1}ab^{n-1} + \binom{n}{n}b^n$$
- This looks scary, but it shouldn't be.
Example
Let's break it down into some steps, using $(2x + 5)^4$ as an example:
- Write out the binomial coefficients (you can do this either using $nCr$ or Pascal's triangle): $$1 + 4 + 6 + 4 + 1$$
- Multiply everything by decreasing powers of the first term, starting at the exponent and ending at $0$: $$1(2x)^4 + 4(2x)^3 + 6(2x)^2 + 4(2x)^1 + 1(2x)^0$$
- Multiply everything by increasing powers of the second term, starting at $0$ and ending at the exponent: $$1(2x)^4(5)^0 + 4(2x)^3(5)^1 + 6(2x)^2(5)^2 + 4(2x)^1(5)^3+1(2x)^0(5)^4$$
- Simplify: $$16x^4+160x^3+600x^2+1000x+625$$
- You can repeat this process with any binomial to produce its expansion.
Finding Terms and Coefficients
To find a specific term in the expansion, we can use the general term formula: $$ \text{The } (r+1)\text{th term} = \binom{n}{r}a^{n-r}b^r $$
Example
- In the expansion of $(2x-3y)^5$, let's find the term containing $x^2y^3$.
- We need $r=3$ because we want $y^3$.
- Term = $$= \binom{5}{3}(2x)^{5-3}(-3y)^3$$ $$= 10 \cdot 2^2 \cdot (-3)^3 \cdot x^2y^3$$ $$= 10 \cdot 4 \cdot (-27) \cdot x^2y^3$$ $$= -1080x^2y^3$$
Common Mistake
Students often forget to include the coefficients of $a$ and $b$ when finding specific terms. Always remember to include these in your calculations!
Exam technique
- Some of the most difficult (and annoying) questions on IB math exams are on the binomial theorem. These look like:
Find the coefficient of $x^8$ in $(2x^4 + 5x^2)^3(\frac1{x^2}+3x)^5$.
- These usually take the form of finding a coefficient of a term in a product of two binomials.
- They look intimidating, but there's a way to break these questions down step-by-step so they are much easier.
Just follow the steps!
- We want to simplify this expression as much as possible.
- So, we can start by factoring out some powers of $x$ to give us a constant term in both binomials. $$(2x^4 + 5x^2)^3(\frac1{x^2}+3x)^5 = [(x^2)(2x^2 + 5)]^3[(\frac1{x^2})(1+3x^3)]^5$$ $$=(x^6)(x^{-10})(2x^2+5)^3(3x^3+1)^5$$
$$=x^{-4}(2x^2+5)^3(3x^3+1)^5$$ - Here's the trick: all the terms in the product $(2x^2+5)^3(3x^3+1)^5$ will be multiplied by $x^{-4}$.
- So, the coefficient of $x^8$ in $x^{-4}(2x^2+5)^3(3x^3+1)^5$ will be the same as the coefficient of $\frac{x^8}{x^{-4}}=x^{12}$ in $(2x^2+5)^3(3x^3+1)^5$.
- So, we've essentially changed the question to "Find the coefficient of $x^{12}$ in $(2x^2+5)^3(3x^3+1)^5$, which is a lot less scary.
- List out the possible powers in the expansions of both binomials. For $(2x^2+5)^3(3x^3+1)^5$, the possible powers for each binomial are:
- $(2x^2+5)^3$: $x^0$, $x^2$, $x^4$, $x^6$
- $(3x^3+1)^5$: $x^0$, $x^3$, $x^6$, $x^9$, $x^{12}$, $x^{15}$
- Multiply the powers of the first binomial expansion with the powers of the second, and find which ones result in the desired product (in this case $x^{12}$.
- For this, only $(x^0)(x^{12})$ and $(x^6)(x^6)$ equal $x^{12}$.
- Find the coefficients for each combination and add them.
- This comes back to the binomial theorem, and for this question, we have: $$[\binom{3}{3}(2x^2)^0(5)^3][\binom{5}{1}(3x^3)^4(1)^1] + [\binom{3}{0}(2x^2)^3(5)^0][\binom{5}{3}(3x^3)^2(1)^3]$$
- Calculate: $$[\binom{3}{3}(2)^0(5)^3][\binom{5}{1}(3)^4(1)^1] + [\binom{3}{0}(2)^3(5)^0][\binom{5}{3}(3)^2(1)^3]=50723$$
- This is very long, but once you have enough practice it becomes easier.
Self review
Find the coefficient of $x^6$ in $(3x + \frac4x)^4(2x^4+6)^3$.
