1The idea in one paragraph

A binomial is a bracket with two terms, such as (a + b) or (2x − 3). Raising it to a power by multiplying out bracket after bracket works but is slow, and it gets slower fast. The binomial theorem does it in one line. Every term of (a + b)ⁿ has the shape ⁿCᵣ aⁿ⁻ʳ bʳ: the powers of a fall from n to 0, the powers of b rise from 0 to n, and the number in front, the binomial coefficient ⁿCᵣ, counts how many ways the term can be made. Those coefficients are the rows of Pascal's triangle, and they are given by a formula that is in the formula booklet. With that one shape you can expand a whole bracket, or pick out just the term you need.

2The pattern, found by hand

Start small and look.

Figure 1 shows (a + b)² as an area: a square of side a + b cut into a², two rectangles ab, and b². The two identical rectangles are where the 2 in 2ab comes from.

Three things hold in every row, and they are the whole structure of the theorem.

• There are n + 1 terms. (a + b)⁴ has five.
• The powers of a go down, the powers of b go up, one step at a time: a⁴, a³b, a²b², ab³, b⁴.
• In every term the two powers add up to n. In 6a²b², 2 + 2 = 4.

What changes from row to row are the numbers in front: 1 1, then 1 2 1, then 1 3 3 1, then 1 4 6 4 1. Section 3 finds them without multiplying.

3Pascal's triangle

Write the coefficients in rows and a pattern appears. Figure 2 shows rows 0 to 7.

Each row starts and ends with 1, and every other entry is the sum of the two entries above it. Row 5 is 1 5 10 10 5 1, so row 6 is 1, 1 + 5, 5 + 10, 10 + 10, 10 + 5, 5 + 1, 1, which is 1 6 15 20 15 6 1. Row n gives the coefficients of (a + b)ⁿ, counting the top row as row 0.

Why the adding rule works: to get (a + b)⁶, multiply (a + b)⁵ by (a + b). The a²b⁴ term of (a + b)⁶ can then arise in two ways, from a × (the ab⁴ term) or from b × (the a²b³ term), so its coefficient is the sum of those two coefficients in row 5.

The triangle is the quickest method for a small n on Paper 1, and it is worth being able to write down the first seven or eight rows in under a minute.

Whose triangle? It carries Pascal's name because he wrote a treatise on it in the seventeenth century. It was not his discovery. The Chinese mathematician Yang Hui set it out in the thirteenth century, and it was known in India and Persia before that. Naming a result after the person who made it famous in Europe, rather than the people who found it, is a pattern in the history of mathematics, and a fair question for TOK: how much of what we call "Pascal's triangle" reflects mathematics, and how much reflects who wrote the history?

4Where the numbers come from: choosing

Pascal's triangle gives the coefficients. Counting explains them, and gives a formula that works for any row without building the rows above it.

Write (a + b)³ as (a + b)(a + b)(a + b). Multiplying it out means choosing either a or b from each of the three brackets and multiplying the choices together. There are 2 × 2 × 2 = 8 ways to choose, and Figure 3 lists all eight, grouped by how many times b was chosen.

Exactly one choice gives a³ (a every time). Three choices give a²b, because b can come from the first, the second or the third bracket. Three give ab² and one gives b³. So the coefficient of a²b is 3 because there are 3 ways to choose which one bracket supplies the b.

The same argument works for any n. The term in aⁿ⁻ʳbʳ comes from choosing b from r of the n brackets (and a from the rest), so its coefficient is the number of ways of choosing r things from n. That number is written ⁿCᵣ (also written as n over r in a tall bracket) and read "n choose r".

To calculate it you need factorial notation: n! (n factorial) means n × (n − 1) × … × 2 × 1, so 4! = 24, and by definition 0! = 1. Then, from the formula booklet:

ⁿCᵣ = n! ÷ (r!(n − r)!)

Worked example 1. Paper 1. Find ⁷C₃ without a calculator.

The cancelling in the second line is the practical trick: ⁿCᵣ is always r numbers counting down from n, divided by r!. So ¹⁰C₂ = (10 × 9) ÷ 2 = 45 without writing any factorials at all.

Two facts save work.

• ⁿC₀ = ⁿCₙ = 1 and ⁿC₁ = ⁿCₙ₋₁ = n, the ends of every row.
• ⁿCᵣ = ⁿCₙ₋ᵣ. Choosing which r brackets give b is the same as choosing which n − r brackets give a. This is why every row of Pascal's triangle is symmetrical.

With technology. On Paper 2 the GDC has an nCr function, and the guide expects you to use it. It also expects you to work backwards from a value using a table.

Worked example 2. Paper 2. Find the values of r for which ⁸Cᵣ = 56.

Enter y = 8 nCr x in the GDC's function table and read the values for x = 0, 1, 2, …, 8. Figure 4 shows them. ⁸Cᵣ = 56 at r = 3 and r = 5.

There are two answers, by symmetry: ⁸C₃ = ⁸C₅. A question that says "find the values" expects both.

When n is the unknown, write out the formula instead. If ⁿC₂ = 105, then n(n − 1) ÷ 2 = 105, so n² − n − 210 = 0, (n − 15)(n + 14) = 0, and n = 15 (n must be a positive whole number, so −14 is rejected).

5The binomial theorem

Put the pieces together. From the formula booklet, for n ∈ ℕ:

The term containing bʳ is the general term:

General term: ⁿCᵣ aⁿ⁻ʳ bʳ. It is term number r + 1, because the first term has r = 0.

Figure 5 labels each piece.

The theorem is stated for a + b, but a and b can be anything: numbers, 2x, −3, x², 1/x. The one rule is that a and b are the whole of each term, sign and coefficient included. In (2x − 3)⁴, a = 2x and b = −3. Put brackets round each when you substitute, so that the sign and the coefficient are raised to the power too.

Worked example 3. Paper 1. Expand (2x − 3)⁴, simplifying each term.

Row 4 of Pascal's triangle is 1 4 6 4 1. Here a = 2x and b = −3.

Check: the signs alternate, as they must when b is negative, because odd powers of −3 are negative. And putting x = 1 into both sides gives (−1)⁴ = 1 and 16 − 96 + 216 − 216 + 81 = 1.

6Finding one term

Most exam questions do not want the whole expansion. They want one term, and writing out all of them wastes time and invites slips. The method is the same every time: write the general term, simplify the powers of x, choose r.

Worked example 4. Paper 1. Find the coefficient of x³ in the expansion of (3 + 2x)⁷.

The coefficient is 22 680. The common error is to forget the 2³ and answer 35 × 81 = 2835: the coefficient inside the bracket is raised to the power too.

Worked example 5. Paper 1. Find the term independent of x in the expansion of (x² − 2/x)⁶.

"Independent of x" means the constant term, the one with x⁰.

Figure 6 shows the power of x in each of the seven terms, dropping by 3 each time, and reaching 0 at r = 4. If 12 − 3r = 0 had no whole-number solution, the expansion would have no constant term, and "there is no term independent of x" would be the answer.

Worked example 6. Paper 1. The coefficient of x² in the expansion of (1 + kx)⁶ is 60. Find the possible values of k.

Both values are valid, because k² hides the sign. If the question had given the coefficient of x³ instead, 20k³ has only one real cube root for each value, and there would be one answer.

Worked example 7. Paper 1. Find the coefficient of x² in the expansion of (1 + 3x)(1 − 2x)⁵.

You only need the first few terms of (1 − 2x)⁵, because the (1 + 3x) can raise the power of x by at most one.

The coefficient is 10. Both pairings must be found: every way of making x² from one term of each bracket.

A use for the first few terms. Because (1 + x)ⁿ = 1 + nx + ⁿC₂x² + … and the later terms are tiny when x is small, the first three terms give a quick estimate. 1.02⁶ = (1 + 0.02)⁶ ≈ 1 + 6(0.02) + 15(0.02)² = 1 + 0.12 + 0.006 = 1.126. The true value is 1.12616…, so three terms are already correct to four significant figures.

At HL, 1.10 extends the theorem to powers that are fractions or negative, where the expansion never ends and this kind of approximation is the main use.