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The Binomial Expansion

Q: What is nr and why does it appear in the expansion?

It is the number of ways to choose r objects from n, equal to n!r!(n-r)!. It appears because the coefficient of each term counts how many of the n brackets contribute a b — exactly a “choose” count.

Q: How do I find one specific coefficient without expanding everything?

Use the general term nra^n-rb^r. Work out which r gives the power of x you want, then evaluate that single term — including any coefficients and signs inside the bracket.

Q: How do the signs work for (a - b)^n?

Treat it as (a + (-b))^n. The factor (-b)^r is positive for even r and negative for odd r, so the terms alternate in sign.

Q: What happens in each term to the powers?

The powers of the two parts always add to n. As r increases, the power of b rises and the power of a falls by the same amount, keeping the total at n.

Expanding $(a + b)^{n}$ by multiplying out is slow and error-prone for large $n$ . The binomial theorem gives every term directly, with coefficients that count combinations — the same $(r n)$ from probability.

30 min Video by TLMaths Sequences and Series

The big picture

Raising a bracket to a high power looks like a chore, but the pattern hiding inside is one of the most elegant in mathematics. The coefficients are the rows of Pascal’s triangle, and each one counts something concrete: how many ways to choose $r$ items from $n$ . That is why the binomial coefficient $(r n)$ appears identically in the binomial distribution in statistics — expansion and counting are the same idea. Mastering the general-term approach lets you pick out a single coefficient from an enormous expansion without writing the rest, and it sets up the Year-2 extension to fractional and negative powers.

A-Level Maths D1-01: Pascal’s Triangle and nCrWatch the walkthrough, then work through the notes below.

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What you'll be able to do

Use Pascal’s triangle and $(r n)$ to find binomial coefficients
Expand $(a + b)^{n}$ for a positive integer $n$
Find a specific term or coefficient using the general term
Handle signs and coefficients inside the bracket carefully

Coefficients: Pascal’s triangle and $\binom{n}{r}$

The coefficients in $(a + b)^{n}$ are the numbers in the $n$ -th row of $Pascal’s triangle$ , where each entry is the sum of the two above it. For small $n$ that is quick; for larger $n$ you compute them directly with the $binomial coefficient$ $(r n)$ (read “ $n$ choose $r$ ”), which counts the ways to choose $r$ objects from $n$ .

(r n) = \frac{n !}{r ! ( n - r )!}

The number of ways to choose

r

from

n

— also the coefficient.

(2 5) = \frac{5 !}{2 ! 3 !} = \frac{120}{2 \times 6} = \frac{120}{12} = 10

Answer

10

Why this works

The coefficient and the “choose” count are literally the same number because expanding $(a + b)^{n}$ asks, for each term, in how many ways you can pick the $b$ ’s from the $n$ brackets — which is exactly $(r n)$ .

The binomial theorem

Putting the coefficients with the right powers gives the full expansion. In each term the powers of $a$ and $b$ add up to $n$ : as the power of $b$ climbs from 0 to $n$ , the power of $a$ falls from $n$ to 0.

(a + b)^{n} = r = 0 \sum n (r n) a^{n - r} b^{r}

Powers of

a

and

b

always sum to

n

1Coefficients (row 4 of Pascal):

1, 4, 6, 4, 1

2Terms with

a = x

b = 2

x^{4} + 4 x^{3} (2) + 6 x^{2} (2)^{2} + 4 x (2)^{3} + (2)^{4}

3Simplify:

x^{4} + 8 x^{3} + 24 x^{2} + 32 x + 16

Answer

x^{4} + 8 x^{3} + 24 x^{2} + 32 x + 16

Tip — When the bracket has a coefficient or a minus, such as $(2 x - 3)^{n}$ , apply the power to the whole term: $(2 x)^{n - r} (- 3)^{r}$ . Odd powers of the negative make the sign alternate.

Picking out a single term

Often you only need one coefficient — say “the coefficient of $x^{5}$ ” — in a large expansion. Rather than expand everything, use the $general term$ $(r n) a^{n - r} b^{r}$ , find the value of $r$ that produces the power you want, and evaluate just that term.

general term = (r n) a^{n - r} b^{r}

Choose

r

to hit the power you need.

1General term:

(r 6) (1)^{6 - r} (2 x)^{r} = (r 6) 2^{r} x^{r}

2We want

x^{3}

, so

r = 3

3Coefficient:

(3 6) 2^{3} = 20 \times 8 = 160

Answer

160

Why this works

The general-term method is a scalpel: it isolates the one term you care about. For “find the coefficient of $x^{k}$ ”, solve for $r$ from the powers first, then compute a single $(r n)$ — never expand the lot.

Think like an examiner

In every term the powers of the two parts must add to $n$ — a fast check that you have not dropped a factor.
Apply the power to coefficients and signs inside the bracket: $(2 x)^{r} = 2^{r} x^{r}$ , and $(- b)^{r}$ alternates sign.
For “coefficient of $x^{k}$ ”, use the general term and solve for $r$ — expanding fully wastes time and invites errors.
Know that $(0 n) = (n n) = 1$ ; these end coefficients are easy marks and quick checks.

Common misconceptions

Writing

(2 x)^{3}

2 x^{3}

inside an expansion.

The power applies to the coefficient too:

(2 x)^{3} = 8 x^{3}

Forgetting the sign pattern in

(a - b)^{n}

(- b)^{r}

is negative for odd

r

, so the signs alternate:

+, -, +, -, \dots

Expanding the whole bracket to find one coefficient.

Use the general term

(r n) a^{n - r} b^{r}

and pick the

r

that gives the required power.

Letting the powers of

a

and

b

fail to add to

n

In every term the exponents sum to

n

; if they do not, a term is wrong.

Binomial essentials

(r n) = \frac{n !}{r ! ( n - r )!}

Binomial coefficient.

(a + b)^{n} = r = 0 \sum n (r n) a^{n - r} b^{r}

The theorem.

(r n) a^{n - r} b^{r}

General term (pick

r

Stretch yourself

Find the coefficient of $x^{2}$ in the expansion of $(3 - 2 x)^{5}$ .

Hint — Use the general term $(r 5) (3)^{5 - r} (- 2 x)^{r}$ . The power of $x$ is $r$ , so set $r = 2$ and evaluate — mind the sign.

Questions students ask

What is

(r n)

and why does it appear in the expansion?

It is the number of ways to choose

r

objects from

n

, equal to

\frac{n !}{r ! ( n - r )!}

. It appears because the coefficient of each term counts how many of the

n

brackets contribute a

b

— exactly a “choose” count.

How do I find one specific coefficient without expanding everything?

Use the general term

(r n) a^{n - r} b^{r}

. Work out which

r

gives the power of

x

you want, then evaluate that single term — including any coefficients and signs inside the bracket.

How do the signs work for

(a - b)^{n}

Treat it as

(a + (- b))^{n}

. The factor

(- b)^{r}

is positive for even

r

and negative for odd

r

, so the terms alternate in sign.

What happens in each term to the powers?

The powers of the two parts always add to

n

. As

r

increases, the power of

b

rises and the power of

a

falls by the same amount, keeping the total at

n

Key takeaways

Coefficients come from Pascal’s triangle, or $(r n) = \frac{n !}{r ! ( n - r )!}$ .
$(a + b)^{n} = \sum (r n) a^{n - r} b^{r}$ ; powers always add to $n$ .
Apply the power to coefficients and signs inside the bracket.
Use the general term to pick out a single coefficient without full expansion.
The same $(r n)$ drives the binomial distribution in statistics.

How this fits the course

Build on

Index Laws

The Binomial Distribution — Statistics (later)soon
Partial Fractions

Leads to

Trigonometric Ratios and Graphs

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