# The Binomial Theorem

> Expand (a + b)ⁿ quickly with Pascal's triangle and combinations, find any term or coefficient, and spot patterns.

From HSC Maths Extension 1 on WhizzNotes, free study notes for NSW students: https://whizznotes.com/mathematics-extension-1/binomial-theorem/. Updated 3 November 2026.
Year 11 and Year 12 boxes go deeper than the Year 10 text around them.

### Story time: The interest shortcut
A bank grows money by 2% a year, so after 10 years each dollar becomes $1.02^{10}$. Before calculators, clerks estimated this by expanding $(1 + 0.02)^{10}$ and keeping only the first few terms:
$$1 + 10(0.02) + 45(0.02)^2 + \dots \approx 1.218$$
The true value is 1.2190, very close. The numbers 1, 10, 45 come from the **binomial theorem**, which expands any power of a two-term bracket without multiplying it out by hand.

**In short:** The binomial theorem expands $(a + b)^n$ without multiplying out: the coefficients come from row $n$ of Pascal's triangle. The general term is $T_{r+1} = \binom{n}{r}a^{n-r}b^r$, which finds any single term, like the term in $x^3$.

## Starting point: what you already know

- Expanding brackets: $(a + b)^2 = a^2 + 2ab + b^2$.
- Index laws: $(2x)^3 = 8x^3$ and $x^a \times x^b = x^{a + b}$.
- Combinations: $\dbinom{n}{r} = \dfrac{n!}{r!\,(n - r)!}$ counts ways to choose $r$ things from $n$ (previous chapter).

## Pascal's triangle

Expand a few powers and look at the numbers in front:

| Power | Expansion | Coefficients |
|---|---|---|
| $(a + b)^1$ | $a + b$ | 1 1 |
| $(a + b)^2$ | $a^2 + 2ab + b^2$ | 1 2 1 |
| $(a + b)^3$ | $a^3 + 3a^2b + 3ab^2 + b^3$ | 1 3 3 1 |
| $(a + b)^4$ | $a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4$ | 1 4 6 4 1 |

Each row starts and ends with 1, and **each inside number is the sum of the two numbers above it**. This is **Pascal's triangle**. The next row is 1 5 10 10 5 1.

## Why combinations appear

$(a + b)^4 = (a + b)(a + b)(a + b)(a + b)$. To get an $a^2b^2$ term, you pick $b$ from 2 of the 4 brackets and $a$ from the rest. There are $\dbinom{4}{2} = 6$ ways to choose those brackets, so the coefficient is 6.

### Key idea: The binomial theorem
$$(a + b)^n = \sum_{r=0}^{n} \binom{n}{r} a^{n-r} b^r = a^n + \binom{n}{1}a^{n-1}b + \binom{n}{2}a^{n-2}b^2 + \dots + b^n$$

Patterns in every expansion:

- There are $n + 1$ terms.
- The power of $a$ goes **down** from $n$ to 0 while the power of $b$ goes **up** from 0 to $n$.
- The powers in each term add to $n$.

### Worked example: Expand (2x − 1)⁴
Here $a = 2x$ and $b = -1$. Coefficients 1, 4, 6, 4, 1:
$$(2x)^4 + 4(2x)^3(-1) + 6(2x)^2(-1)^2 + 4(2x)(-1)^3 + (-1)^4$$
$$= 16x^4 - 32x^3 + 24x^2 - 8x + 1$$

### Watch out: Brackets around each part
Keep $(2x)$ and $(-1)$ in brackets when you raise them to powers. $(2x)^3 = 8x^3$, not $2x^3$.

## The general term

The term with $b^r$ is

$$T_{r+1} = \binom{n}{r} a^{n-r} b^r$$

It is called $T_{r+1}$ because the first term has $r = 0$.

### Worked example: Find the coefficient of x³ in (x + 3)⁷
We need $x^3$, so $a = x$ has power $7 - r = 3$, giving $r = 4$.
$$\binom{7}{4} x^3 \cdot 3^4 = 35 \times 81\, x^3 = 2835x^3$$
The coefficient is 2835.

### Calculator tip: Negative terms need brackets
When a term is negative, put it in brackets before raising it to a power: $(-2)^5$ is `( -2 ) ^ 5` = $-32$. For a coefficient like $\binom{7}{3}(2)^4(-1)^3$, type it in one line with `nCr` and brackets, and check the sign at the end.

**Mind map**

```text
Binomial theorem
  Pascal's triangle
    Add the two above
    Row n gives the coefficients
  Formula
    Sum of nCr a^(n-r) b^r
    n + 1 terms
    Powers add to n
  General term
    T(r+1) = nCr a^(n-r) b^r
    Find a coefficient
    Find the constant term
  Uses
    Approximations
    Identities with nCr
```

### Year 11: The term independent of x
Find the constant term in $\left(x^2 + \dfrac{1}{x}\right)^6$.

General term: $\dbinom{6}{r}(x^2)^{6-r}\left(\dfrac{1}{x}\right)^r = \dbinom{6}{r}x^{12 - 2r - r} = \dbinom{6}{r}x^{12 - 3r}$.

Constant needs $12 - 3r = 0$, so $r = 4$. The term is $\dbinom{6}{4} = 15$.

### Year 11: Pascal's rule and coefficient identities
**Pascal's rule:** $\dbinom{n}{r} = \dbinom{n - 1}{r - 1} + \dbinom{n - 1}{r}$. This is the "add the two above" rule written with combinations.

Put $a = b = 1$ in the theorem:
$$\binom{n}{0} + \binom{n}{1} + \dots + \binom{n}{n} = 2^n$$
This makes sense: each of $n$ items is either in or out of a subset, so there are $2^n$ subsets in total.

**Sum of the coefficients.** Put $x = 1$. The coefficients of $(2x - 1)^6$ add to $(2 - 1)^6 = 1$.

**Two brackets multiplied.** For the coefficient of $x^2$ in $(1 + x)^3(1 - x)^4$, list the low terms of each and pair the powers that add to 2: $(1 + 3x + 3x^2)(1 - 4x + 6x^2)$ gives $6 - 12 + 3 = -3$.

**Finding $n$.** If the coefficient of $x^2$ in $(1 + x)^n$ is 28, then $\binom{n}{2} = \dfrac{n(n - 1)}{2} = 28$, so $n = 8$.

**When does a constant term exist?** In $\left(x^3 + \frac{1}{x}\right)^n$ the general term has $x^{3(n - r) - r} = x^{3n - 4r}$. A constant term needs $r = \frac{3n}{4}$ to be a whole number, so $n$ must be a multiple of 4.

**The greatest coefficient.** Compare neighbours. In $(1 + 3x)^8$ the coefficient of $x^r$ is $\binom{8}{r}3^r$, and the ratio of the next coefficient to this one is $\dfrac{3(8 - r)}{r + 1}$. It is bigger than 1 while $24 - 3r > r + 1$, that is $r < 5.75$, so the coefficients grow up to $r = 6$ and then shrink: $x^6$ has the greatest coefficient.

### Year 12: Towards the binomial distribution
If a trial succeeds with probability $p$ and fails with $q = 1 - p$, then $(q + p)^n = 1$. Each term $\dbinom{n}{k}q^{n-k}p^k$ is the probability of exactly $k$ successes. The binomial theorem becomes the binomial distribution in Year 12.

**Identities from calculus.** Differentiate $(1 + x)^n = \sum\binom{n}{r}x^r$ to get $n(1 + x)^{n-1} = \sum r\binom{n}{r}x^{r-1}$, then put $x = 1$: $\sum r\binom{n}{r} = n2^{n-1}$. Integrating from 0 to 1 instead gives $\sum\dfrac{1}{r + 1}\binom{n}{r} = \dfrac{2^{n+1} - 1}{n + 1}$.

## Quick summary

- Pascal's triangle: each number is the sum of the two above. Row $n$ gives the coefficients of $(a + b)^n$.
- $(a + b)^n = \displaystyle\sum_{r=0}^{n} \binom{n}{r}a^{n-r}b^r$, with $n + 1$ terms.
- General term $T_{r+1} = \dbinom{n}{r}a^{n-r}b^r$. Set the power of $x$ to what you need and solve for $r$.
- The sum of row $n$ of Pascal's triangle is $2^n$.

## Quiz

Try these questions from the chapter quiz.

1. How many terms are in the expansion of $(a + b)^9$?
   - A. 9
   - B. 10
   - C. 8
   - D. 18
   - Answer: B. 10. There are $n + 1$ terms.

2. Which row of Pascal's triangle gives $(a + b)^5$?
   - A. 1 4 6 4 1
   - B. 1 5 10 10 5 1
   - C. 1 5 10 5 1
   - D. 1 6 15 20 15 6 1
   - Answer: B. 1 5 10 10 5 1. Add pairs from the row 1 4 6 4 1.

3. Expand $(x + 1)^3$.
   - A. $x^3 + 3x^2 + 3x + 1$
   - B. $x^3 + 1$
   - C. $x^3 + 3x + 1$
   - D. $x^3 + x^2 + x + 1$
   - Answer: A. $x^3 + 3x^2 + 3x + 1$. Coefficients 1, 3, 3, 1.

The whole chapter quiz is on the page, for signed-in students: https://whizznotes.com/mathematics-extension-1/binomial-theorem/#quiz